Tag Info

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)+KObserve 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: $\int{x \sqrt{2x+1}dx }=\int{\frac{t^2-1}{2}t^2dt}=\frac{1}{2}\int(t^4-t^2)dt$

Top 50 recent answers are included