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0

Of course that's correct. The reason for doing that is because then you get a integral which can easily be integrated using the power rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + c$. Also when you perform the substitution you need everything in terms of $u$ so you may manipulate the substitution expression however you want to in order to get the integral ...


0

I followed the same approach as I used in an answer to another question, and expanded your integral in multiple polylogarithms of weight 4, then used some patterns in their values of weight 3 to guess terms that might appear in the integral. Then I used an integer relation algorithm to express your integral in terms of logs, zeta functions and polylogarithms ...


0

$\def\Li{\,\mathrm{Li}}$I followed the technique suggested by Julian Rosen in his answer, and decomposed your integral (and your other integral) as a linear combination of multiple polylogarithms: $$\textstyle -\frac12\log2\log3 \Li_2({\frac23}) + \frac12\log3\Li_{2,1}({\frac23,\frac34}) + \frac12\log2\Li_{2,1}({\frac23,1}) \\\textstyle - ...


0

Expanding @Lucian's comment, if $a=\frac{p}{q}\in\mathbb{Q}$, we have: $$\begin{eqnarray*}I(a)&=&q\int_{0}^{+\infty}\left(1-\frac{\tanh px}{\tanh ...


0

As answered by Karl $$\int t^{\kappa } \exp{\left(-\rho t^{\alpha\kappa + 1}\right)} \, dt=-\frac{t^{\kappa +1} \left(\rho t^{\alpha \kappa +1}\right)^{-\frac{\kappa +1}{\alpha \kappa +1}} \Gamma \left(\frac{\kappa +1}{\alpha \kappa +1},t^{\alpha \kappa +1} \rho \right)}{\alpha \kappa +1}$$ where appears the incomplete gamma function. This can ...


0

Did you try the substitution $u=t^{\alpha\kappa+1}$? As far as I see, you get then something like $k\cdot u^{\beta}\exp(-\rho u)$ ($k,\beta$ constants) as integrand, the integral is then similar to the incomplete Gamma function. https://en.wikipedia.org/wiki/Incomplete_gamma_function Hope that it helps.


2

Just because: $$\forall n\in\mathbb{N}, \quad \int_{0}^{+\infty} x^n\,e^{-x}\,dx = \Gamma(n+1)=n!.$$


1

$$ 7\int \frac{1}{x^2+x\sqrt{x}}\ dx $$ Let $u=\sqrt{x}$, then $$ du = \frac{1}{2\sqrt{x}}dx $$ So now we have $$ 14\int \frac{u}{u^4+u^3}\ du=14\int \frac{1}{u^3+u^2}\ du $$ The next step is to simplify the integrand via partial fraction decomposition $$ 14\int \frac{1}{u^2}du+14\int \frac{1}{u+1}du-14\int \frac{1}{u}du $$ Can you take it from here?


1

As answered by ASKASK, using $x=u^2$, $$\int\frac{dx}{x^2+x\sqrt x}=2\int\frac{du}{u^3+u^2}$$ Now, use partial fraction decomposition to get $$\frac{1}{u^3+u^2}=\frac{1}{u^2(u+1)}=\frac{2}{u^2}+\frac{2}{u+1}-\frac{2}{u}$$and then integrate each piece. When done, go back to $x$ since $u=\sqrt x$. I am sure that you can take from here.


-1

hint: $$u=\sqrt{x}$$ $$du=\frac1{2u}dx$$ $$2udu=dx$$ $$7\int\frac{1}{u^4+u^3}*2udu$$ $$7\int\frac{2u}{u^4+u^3}du$$ And there's your rational function


1

Your $\int x dx$ should be $\int \frac12 x dx$. Then your final answer will be $\frac12 x^2(\ln x-\frac12)$


2

Divide $2y^2+1$ by $-y+3$. We get $-2y-6+\frac{19}{-y+3}$. Now the integration should be straightforward.


0

Try the substitution: $\sqrt{x}=t$


2

Make the substitution $y = \sqrt{x}$. Then $x=y^2$ so that $dx = 2y\,dy$, and $$ \int \frac{dx}{x^2+x\sqrt{x}} = \int \frac{2y\,dy}{y^4+y^3} = 2\int \frac{dy}{y^2(y+1)}. $$ You should be able to carry on from here?


0

Hint: $\int\frac{5x^2+2x-5}{x^3-x}dx=\int\frac{5(x^2-1)+2x}{x(x^2-1)}dx=\int\frac{5(x^2-1)}{x(x^2-1)}dx+\int\frac{2}{x^2-1}dx=\int\frac{5}{x}dx+\int\frac{2}{x^2-1}dx=$ $\ 5\int\frac{1}{x}dx+2\int\frac{1}{x^2-1}dx$ The first is easy now, for the second you have to manipulate a little: ...


1

A start: Use partial fractions. Find constants $A,B,C$ such that your function is equal to $\displaystyle \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$.


1

Although the $\sqrt{1-x^2}$ does suggest a trig substitution, the positive power of $x$ on the outside makes it easier to do this integral by parts: Let $u=x^2$ and $dv = x\sqrt{1-x^2}dx$ $$ \begin{array}{cc} u=x^2 & v = - \frac{1}{3}(1-x^2)^{3/2} \\ du = 2x\,dx & dv = x\sqrt{1-x^2}dx \end{array} $$Then $$ \int_0^1 u\, dv = \left.uv\right|_0^1 - ...


0

$$14\int_0^1x^3\sqrt{1-x^2}dx=14\int_0^1x^2\sqrt{1-x^2}xdx=14\int_1^0(1-t)\sqrt t\frac{dt}{-2}=$$ $$=7\int_0^1(1-t)\sqrt tdt=7\int_0^1t^{1/2}dt-7\int_0^1t^{3/2}dt=...$$ $1-x^2=t,-2xdx=dt,xdx=\frac{dt}{-2},x^2=1-t$


0

You can show that for any polynomial $f(x)$ there exists an $a>0$ such that $$\left|\dfrac{f(x)}{e^x}\right| < \dfrac{1}{x^2}$$ for all $x>a$. In fact it's easy. If $f(x)=a_nx^n+\cdots + a_1x + a_0$, then for some $c>0$ there is an interval $(c,\infty)$ with $|a_nx^n| \ge |a_ix^i|$, $i=0,...,n$. So $|f(x)|\le (n+1)|a_n|x^n$ for all $x\in ...


0

Since $fg$ is a polynomial there exist a $M$ s.t. when $x > M$ we have $$|fg| \leq Ax^{k}$$ for a positive constant $A$ and an integer $k$. This follows from writing $$fg = \sum_{i=0}^na_n x^n = a_nx^n\left(1 + \frac{a_{n-1}}{a_n x} + \ldots + \frac{a_0}{a_n x^n}\right)$$ The terms in the brackets convergest to $1$ when $x\to \infty$ so we can find ...


0

If you mean $\frac{f(x)*g(x)}{e^x }$ then this should be the same as $\frac{h(x)}{e^x }$ since $f(x)*g(x) \in R[x]$. Then with l'Hopital's rule you should see that if $h(x)$ has power c, after doing l'Hopital's rule c times you have $\frac{1}{e^x}$. So $\int{\frac{1}{e^x}{dx}}$ = $\int{{e^{-x}}{dx}}$ = $-{e^{-x}}$ which will go to 0 as x->$\infty$. Not ...


1

Since $fg$ is a polynomial, it suffice to prove that $\displaystyle \int_0^\infty \frac{F}{e^x} dx$ is convergent for a single polynomial $F$. Further, note that if $F = \displaystyle\sum_{i=0}^k (a_i x^i)$ is a degree $k$ polynomial, then $\displaystyle \int_0^\infty \frac{F}{e^x} dx = \sum_{i=0}^k a_i \int_0^\infty \frac{x^i}{e^x} dx$. Therefore it ...


1

HINT: $$\lim_{x -> \infty} \frac{ f g\ (x) /e^x } { e^{-x/2}} = \lim_{x->\infty}\frac{f g\ (x)}{e^{x/2}}=0$$ with l'Hospital's rule.


4

We start from your $$\int \frac{1}{a}\cdot \frac{1}{1+(x/a)^2}\,dx.$$ Let $u=\frac{x}{a}$. Then $du=\frac{1}{a}\,dx$, so $dx=a\,du$. Our integral becomes $$\int \frac{1}{a}\cdot \frac{1}{1+u^2}a\,du.$$ Note the cancellation of the $a$'s, and integrate. We get $$\arctan(u)+C,\quad\text{which is}\quad \arctan(x/a)+C.$$ Remarks: $1.$ In the first iteration of ...


1

The question is poorly stated as the independent variable is not defined.


3

Setting $x=a\tan\theta,\dfrac xa=\tan\theta,\theta=\arctan\dfrac xa$ $$\int\frac1{1+(x/a)^2}dx=\int a\ d\theta=a\arctan(x/a)+K$$ Setting $a=1,$ we find $$\int\frac{dx}{1+(x)^2}=\arctan(x)+K$$ Observe the two $x$-s within parentheses


0

Since the function $\phi(t)$ is defined on $-L=-\pi<t<\pi=L$ , the Fourier series can be expressed as shown below : The numerical tests of the formula are well consistent with a good accuracy. On the figure below, small values of $m$ are taken in order to make clear the deviations in case of series limited to not enough terms.


3

Ms. Chris's sis asked me exactly same question a few days ago in chatroom & I could answer it. Here is my answer. Let $I$ be the integral. Using magic substitution $2t=1+x$ we get \begin{align} I&=\int_{\frac{1}{2}}^1 \frac{\log(2t)\log(2-2t)}{t}dt\\ &=\int_{\frac{1}{2}}^1 \frac{\log t\log(1-t)}{t}dt+\ln2\int_{\frac{1}{2}}^1 ...


0

Let's assume that you want to measure the amount of the slope between two points on the function $f(x)$, which are $f(x)$ and $f(x+h)$. This is defined as: $$M(h)=\dfrac{f(x+h)-f(x)}{h}$$. $M$ is actually the slope of the secant line which joins the points $(x,f(x))$ and $(x+h,f(x+h))$. Notice that, when we determine a fixed $x$ value, $M(h)$ is a function ...


1

Using Maple I am obtaining $$1+\frac{\pi }{16}{\ _4F_3(1,1,1,3/2;\,2,2,2;\,1)}+\frac{\sqrt {\pi }}{8} G^{4, 1}_{4, 4}\left(-1\, \Big\vert\,^{1, 5/2, 5/2, 5/2}_{2, 3/2, 3/2, 1}\right) $$ and a numerical approximation is $$1.3913063720392030337$$


0

The derivative function says how fast the original function is changing at each point. If $f(t)$ is the position of a particle or a rocket ship at each time $t$, then the derivative $f'(t)$ is the speed of the particle or the rocket ship at time $t$. Consider as an example $f(t) = -5t^2 + 20t$. Suppose this describes the height of a rocket above the ...


0

Since I don't know how to provide sketches in this environment, I will try to give a verbal description of D and E: D essentially consists of 3 pieces: 1) Its top is the portion of the paraboloid $z=4-x^2-y^2$ which lies above the xy-plane. (Notice that the intersection of the paraboloid with the xy-plane is the circle $x^2+y^2=4$.) 2) Its bottom is ...


2

The domain $D$ looks roughly like a right triangle $ABC$ with the right angle $B$ at $(-1,1)$, $A$ at $(-1,-1)$, $C$ at $(1,1)$ and a curve $y=x^3$ instead of a straight line from $A$ to $C$. Since the curve does not do anything tricky (one value of $x$ maps to one value of $y$ and vice-versa) you can do this as a single integral, integrating either $x$ or ...


1

Hint: Break this up into to integrals, one from $-\infty$ to $0$ where $|x|=-x$ and one from $0$ to $2$ where $|x|=x$.


3

Note that $$\begin{align}\int_{-\infty}^{2}0.1\ e^{-0.2|x|}\;\mathrm{d}x&=\int_{-\infty}^{0}0.1\ e^{-0.2(-x)}\;\mathrm{d}x+\int_{0}^{2}0.1\ e^{-0.2x}\;\mathrm{d}x\\&=0.1\int_{-\infty}^{0}e^{0.2x}\;\mathrm{d}x+0.1\int_{0}^{2}e^{-0.2x}\;\mathrm{d}x.\end{align}$$ Here, you can use $$\int e^{ax}\;\mathrm{d}x=\frac 1a e^{ax}+C$$ for $a\not =0$.


0

$\frac{67}{90}$ doesn't look correct. Here is what wolfram computes


3

It seems that you already realize that the for $D$, $D = \{x^3 \leq y \leq 1\} \cap \{-1 \leq x\}$ is the same as $D = \{x^3 \leq y \leq 1\} \cap \{-1 \leq x \leq 1\}$. So, for a function $f(x,y)$, you should have $$ \iint_D f(x,y) \, dxdy = \int_{-1}^1 \int_{x^3}^1 f(x,y)\,dxdy. $$ However, you seem to have split up your function $f(x,y)$ over different ...


3

According to a CAS, $$I = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{t} \right)}{2 \, \sqrt{t} \, \sqrt{1-t}} \,dt =\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right )+\frac{\pi ^3}{48}-\frac{1}{4} \pi \log ^2(2)$$ Enjoy !


0

In a probability course, for the integral there is basically nothing to do. By symmetry, our integral is $$\int_0^\infty (0.2)e^{-0.2 x}\,dx.$$ Since $(0.2)e^{-0.2 x}$ in the interval $(0,\infty)$ is the density function of a well-known distribution (exponential, parameter $0.2$), the integral is $1$.


0

A contrieved approach: First derive the exponential to understand how to handle the absolute value: $$e^{-a|x|}=-a\ sgn(x)e^{-a|x|}.$$ Then the antiderivative could be $$\int e^{-a|x|}dx=-\frac{sgn(x)}ae^{-a|x|},$$ but this is not correct because there is a discontinuity at $0$. We can fix it by adding a sign function (having a null derivative) with a ...


0

To prove it is a density function: $$\int_{-\infty}^{\infty} \frac{1}{10} e^{-|x|/5}dx = 2\int_{0}^{\infty} \frac{1}{10} e^{-x/5}dx = -e^{x/5}|_{0}^{\infty} = 1$$ For the second part of the question, the statement isn't clear but presumably "errors" means that this is a distribution of errors, in which case the answer is $\frac{1}{2}$ since $f$ is symmetric ...


0

$$\frac {1}{10} \int _{-\infty}^{\infty} e^{-0.2 (abs(x)} dx$$ But $e^{-.2 abs(x)}$ is an even function and the interval $(-\infty, \infty)$ is symmetric about $0$, $$\int _{-\infty}^{\infty} e^{-0.2 abs(x)} dx = 2 \int_0^{\infty} e^{-.2 abs(x)} dx$$ So now we have $$ \frac15 \int_0^{\infty} e^{-x/5} dx$$ Substitute $u=-\frac {x}{5}$ and $du = ...


2

$$\int _{-\infty }^{\infty }\! 0.1\,{{\rm e}^{- 0.2\, \left| x \right| } }{dx}=\int _{-\infty }^{0}\! 0.1\,{{\rm e}^{ 0.2\,x}}{dx}+\int _{0}^{ \infty }\! 0.1\,{{\rm e}^{- 0.2\,x}}{dx} $$ $$\int _{-\infty }^{\infty }\! 0.1\,{{\rm e}^{- 0.2\, \left| x \right| } }{dx}={\frac {1}{2}}\int _{-\infty }^{0}\! 0.2\,{{\rm e}^{ 0.2\,x}}{dx}+{\frac {1}{2}}\int _{0}^{ ...


1

Using Maple I am obtaining directly $$4\,R \left( {R}^{2}-{\epsilon}^{2} \right) {\it EllipticE} \left( { \frac {\epsilon}{R}} \right)$$


2

I tried first the antiderivative $$\int\left(R^{2}-\epsilon^{2}\right)\sqrt{R^{2}-\epsilon^{2}\sin^{2}\left(\theta\right)}d\theta=\frac{R^2 \left(R^2-\epsilon ^2\right) \sqrt{\frac{2 R^2+\epsilon ^2 \cos (2 \theta )-\epsilon ^2}{R^2}} E\left(\theta \left|\frac{\epsilon ^2}{R^2}\right.\right)}{\sqrt{2 R^2+\epsilon ^2 \cos (2 \theta )-\epsilon ^2}}$$ So ...


1

Consider $z=-r$ and $z=-\sqrt{4-r^2}$. You have that if $r>0$, so $-r=-\sqrt{4-r^2}\iff r^2=4-r^2\iff r=\sqrt 2$. Then, $$D=\left\{(r\cos\theta,r\sin\theta, z)\mid \theta\in[0,2\pi[, r\in[0,\sqrt 2], z\in[0,-\sqrt 2]\right\}$$$$\cup\left\{(r\cos\theta\sin\varphi, r\sin\theta\sin\varphi, r\cos\varphi)\ \big|\ r\in[0,2], ...


2

The lower cone begins at $\phi=\dfrac{3\pi}{4}$ and runs all the way down to $\phi=\pi$. If the starting point for $\phi$ isn't obvious, then you can find it as follows Let $x=r\sin\theta\cos\phi, y=r\sin\theta\sin\phi, z=r\cos\theta$. Subbing this into the cone equation, squaring and rearranging you obtain $$r^2\cos^2\theta = ...


1

Put u=2x+1. Then $\frac{$d$u}{$d$x}=2 \implies $d$u=2$d$x$. Then, $\int x \sqrt{2x+1}= 0.5 \int 2x \sqrt{2x+1}$d$x=0.5 \int (\frac{u-1}{2}) \sqrt u $ d$u$ Which is easy to integrate.


4

It looks like to integrate the product, you found the product of the integration of each factor. We can't do that! If $\int f(x) \,dx = F(x)$ and $\int g(x)\,dx = G(x)$ $$\int f(x)\cdot g(x) \,dx \neq F(x)G(x) + C$$ Let's start over. Note that since the integrand is defined only for $2x+1\geq0$, we put $$\underbrace{u^2 = 2x+1}_{u = \sqrt{2x+1}} ...


1

The integral of a product is not the product of the integrals. Try to put $\ \sqrt{2x+1}=t$ so you have: $\ 2x+1=t^2 \implies x=\frac{t^2-1}{2}$, $dx=tdt$ and now it should be easier: $\int{x \sqrt{2x+1}dx }=\int{\frac{t^2-1}{2}t^2dt}=\frac{1}{2}\int(t^4-t^2)dt$



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