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1

That notation is used in the classic textbook Elementary Differential Equations by William E. Boyce and Richard C. DiPrima, at least in the third edition (1976), which is the one I have. Quoting from p. 11 (beginning of Chapter 2):      The simplest type of first order differential equation occurs when $f$ depends only on $x$. ...


3

Yes, there is a way without partial fraction. This might also be in your interest. Start off with the substitution: $$x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$$ This substitution produces a nice cancelation in the denominator, since: $$(1+t)^3+(1-t)^3=1+3t +3t^2+t^3 +1-3t+3t^2 -t^3=2(1+3t^2)$$ $$\int \frac{1}{1+x^3}dx=-\int\frac{1}{\frac{(1+t)^...


3

In general, that integral diverges. That will be the case if, say, $x_0=0$, $f(x)=x-1$ and $g(x)=x$. So, yes, some extra hypothesis is required here. Since (assuming that the definition of $h$ makes sense) $h'(x)=\dfrac{f(x)}{g(x)}$, and since this quotient is smaller than $0$ on $[x_0,x_1)$ and greater than $0$ on $(x_1,\infty)$, you are right.


2

Let $$x=y^{25}\implies dx=25\,y^{24}\,dy$$ $$\int\frac{dx}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}=25\int\frac{ y^{15}}{y^{32}+1}\,dy=\frac {25}{16}\int\frac{16\, y^{15}}{y^{32}+1}\,dy=\frac {25}{16}\int\frac{ \left(y^{16}\right)'}{\left(y^{16}\right)^2+1}$$


6

This is a cute $u$-substitution problem. The crux of the problem is dividing out by the right factor of $x$. Notice that if we factor out one copy of $x$ we obtain $$ \int \frac{1}{x\cdot x^{\frac{16}{25}} + x^{\frac{9}{25}} }dx \;\; =\;\; \int \frac{1}{x \left (x^{\frac{16}{25}} + x^{\frac{-16}{25}} \right )}dx. $$ Now let $u = x^{\frac{16}{25}}$, and ...


2

You can apply any substitution to any integral; you may have seen this particular one recommended for integrals of the form you mention, but that doesn't mean you can't use it elsewhere as well. Of course, to use a substitution, you need to satisfy the hypotheses of whatever theorem you have that allows substitutions, but in practice, we often proceed by ...


0

Slightly different approach to Robert Israel's. Using $$\frac{1}{1+x}=1-x+x^2-x^3+x^4-\cdots,\tag{1}$$ where $|x|<1$ you can substitute $x^n$ for $x$ to obtain $$\int\frac{1}{1+x^n}dx=\int 1-x^n+x^{2n}-x^{3n}+x^{4n}-\cdots dx$$ So we end up with $$\int\sum_{k=0}^\infty (-1)^kx^{kn}dx=\sum_{k=0}^\infty(-1)^k\frac{x^{kn+1}}{kn+1}+c=\frac{x}{n} \Phi \left(-...


2

$$I=\int\frac{dx}{7+5\cos x}=\int \frac{\sec^2\left(\frac{x}{2}\right)}{2\left(\tan^2\left(\frac{x}{2}\right)+6\right)}~dx$$ putting $~u=\frac{1}{\sqrt{6}}~\tan\left(\frac{x}{2}\right)~$, we have $$I=\frac{1}{\sqrt{6}}~\int \frac{du}{u^2+1}$$ $$=\frac{1}{\sqrt{6}}~\tan^{-1}u+C$$ $$=\frac{1}{\sqrt{6}}~\tan^{-1}\left(\frac{1}{\sqrt{6}}~\tan\left(\frac{x}{2}\...


4

Integrals of this form are normally easily solved using the Tangent half-angle substitution: $$t=\tan\frac x2$$ $$dx=\frac 2{1+t^2}dt$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ so this would give you: $$\int\frac{dx}{7+5\cos(x)}dx=\int\frac{1}{7+5\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}dt=\int\frac{2}{7(1+t^2)+5(1-t^2)}dt$$ Now you should be able to solve it


1

Use the so-called Weierstrass substitution: $$\cos(x)=\frac{1-t^2}{1+t^2}$$ $$dx=\frac{2dt}{1+t^2}$$


6

Compute it via the Riemann sums over the subdivision $x_k=aq^k$ with $x_N=aq^N=b$. Then \begin{align} \int_a^b\ln(x)dx=\lim_{N\to\infty}\sum_{n=0}^{N-1}(\ln(a)+n\ln(q))aq^n(q-1) \end{align} and using the sum formulas for geometric sums $\sum_{n=0}^{N-1}q^n=\frac{q^N-1}{q-1}$ and $\sum_{n=0}^{N-1}nq^n=\frac{q-Nq^N+(N-1)q^{N+1}}{(q-1)^2}$ we get \begin{align} ...


1

$$\dfrac{d(x^m\ln x)}{dx}=x^{m-1}+mx^{m-1}\ln x$$ $$\implies x^m\ln x=\int x^{m-1}\ dx+m\int(x^{m-1}\ln x)dx$$ Set $m=1$


2

Integrate the power series: \begin{align} \int \log x\, dx &= \int \sum_{i=1}^{\infty} (-1)^{i+1} \frac{(x-1)^i}{i} \\ &= \sum_{i=1}^{\infty} (-1)^{i+1} \int \frac{(x-1)^i}{i} \\ &= \sum_{i=1}^{\infty} (-1)^{i+1} \frac{(x-1)^{i+1}}{i(i+1)} \\ &= (x-1)\sum_{i=1}^{\infty} (-1)^{i+1} \frac{(x-1)^i}{i(i+1)} \\ &= (x-1)\sum_{i=1}^{\infty} (-1)...


17

In order to prove $\int \log x \, \mathrm{d}x = x \log x - x + \mathsf{C}$, it suffices to show that $$ \int_{1}^{x} \log t \, \mathrm{d}t = x \log x - x + 1. $$ We establish this with different methods, avoiding integration by parts technique and 'guessing the antiderivative' strategy. Method 1. Assume $a \geq 1$. Then by Fubini's theorem, \begin{align*}...


4

Graph $y=\ln x$. The area under the curve from $x=1$ to $x=x_0$ is an anti-derivative of $\ln x_0$. Make the rectangle with base on the $x$-axis from $x=0$ to $x=x_0$ and height equal to $\ln x_0$. The area you want is $x_0 \ln x_0$ minus the area above the curve and inside the rectangle. That is: $$\int_1^{x_0} \ln x \; dx = x_0\ln x_0 - \int_0^{\ln ...


2

Hint: Try writing the Taylor series expansion of $\ln(x)$ at $x=1$ (this involves two cases), integrate it and then compare it with the Taylor series expansion of $x\ln(x)-x$. PS: But to think of this tricky solution, it is very important to know the answer already (which really doesn't makes sense). Therefore, its always better to use a hammer for a nail ...


3

Recall that $\ln x$ is the inverse of the function $e^x$. We have the following identity for the integrals of inverse functions: $$\int f^{-1}(x)\,dx=xf^{-1}(x)-F\circ f^{-1}(x)+C,$$ where $F$ is the antiderivative of $f$, not $f^{-1}$. Using this, we obtain the integral of $\ln x$ very easily: $$\int\ln x\,dx=x\ln x-e^{\ln x}+C=x\ln x-x+C$$


1

Just calculate $$\left(x\ln{x}-x\right)'.$$


5

The following development may seem unmotivated, but it does work. Let $$ v := e^x, \;\; w := 1 + x + v, \;\; w' = 1 + v = w - x. $$ We are trying to integrate $$ u := \frac{x^2+x}{(e^x+x+1)^2} $$ Notice the equalities $$ u = \frac{x (1 + x)}{w^2} = \frac{x (w - v)}{w^2} = \frac{x}{w} - \frac{x v}{w^2} = 1 - \frac{w - x}w - \frac{x v}{w^2} = 1 - \frac{w'...


2

$\newcommand{\d}{\mathrm{d}}$ I don't think there are any simple ways to evaluate this integral that aren't ad hoc. Write the expression to the right of the integral sign as $$\omega=w^2(x^2 + x)\d x$$ where $w^{-1}=\mathrm{e}^x + x + 1$. Write $$\alpha=w^2x\d x - w \d x - \d w\text{.}$$ Note that the differential equation that $w$ satisfies is $\alpha = 0$...


0

Avoiding substitution (as request by another question) \begin{eqnarray*} \int \frac{dx}{1+e^x} &=& \int \left( 1-\frac{e^x}{1+e^x} \right) dx \\ &=& \int \left( 1+\sum_{n=1}^{\infty} (-1)^n e^{nx} \right) dx \\ &=& x+\sum_{n=1}^{\infty} \frac{(-1)^n}{n} e^{nx} +C \\ &=& x- \ln(1+ e^x) +C. \end{eqnarray*}


2

$\arcsin{\sqrt{x}}+\arccos{\sqrt{x}}=\frac{\pi}{2}$ and $$\int\arcsin{\sqrt{x}}=x\arcsin{\sqrt{x}}-\int\frac{x}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt{x}}dx$$ Can you end it now?


0

The suggested strategy you found works because of two observations. First, note that the two factors in the denominator of the integrand you start with appears as separate denominators in the second line. If you don't see why the middle equation holds, try to put the two terms on a common denominator: $x^2\sqrt{x^4+x^2+1}$. Second, the two integrands in ...


4

let $$I=\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}dx=\int \frac{x-x^{-3}}{\sqrt{x^2+x^{-2}+1}}dx=\frac{1}{2}\int \frac{dt}{\sqrt t}$$ $$=\sqrt{x^2+x^{-2}+1}+C$$ In the last one we use $x^2+x^{-2}+1=t.$


2

This is a super clever integral. We factor out an $x$ from the square root to get $$I=\int\frac{x^4-1}{x^3\sqrt{x^2+1+1/x^2}}dx$$ If we add and subtract $1$ from within the square root, we can write this as $$I=\int\frac{x^4-1}{x^3\sqrt{\left(x+\frac{1}{x}\right)^2-1}}dx$$ We will let $x+1/x=\sec\theta$ for our substitution, so $(1-1/x^2)dx=\sec\theta\...


8

The observation of @YiFan is right. You actually have $$ \eqalign{ & y' = 3a\left( {\left( {x + {b \over {3a}}} \right)^{\,2} + {{k^{\,2} } \over {\left( {3a} \right)^{\,2} }}} \right)\quad \Rightarrow \cr & \Rightarrow \quad dx = {{dy} \over {3a\left( {\left( {x + {b \over {3a}}} \right)^{\,2} + {{k^{\,2} } \over {\left( {3a} \right)^{\,...


4

Here's a different approach, actually it doesn't use partial fractions, or maybe we can call it pseudo-partial fraction. Start by letting $n=\sqrt[3]{5} x$ then: $$I=\int\frac{n}{n^3+5}dn=\frac{1}{\sqrt[3]{5}}\int\frac{x}{x^3+1}dx$$ Now we will substitute $x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$. The reason behind this substitution is that ...


3

$\textbf{Hint:}$ notice that $$x^3+5 = (x+5^{1/3})(x^2-5^{1/3}x+5^{2/3})$$ So, $$\frac{x}{x^3+5} = \frac{A}{x+5^{1/3}}+\frac{Bx+C}{x^2-5^{1/3}x+5^{2/3}}$$ for some constants $A$, $B$ and $C$.


1

If $u=e^x$ then $du=e^x dx$ and $du= u dx$: this in turn implies $dx= du/u$ so that the integrand gets $$ \frac{1} {(u+1)u} = \frac{1}{u} - \frac{1}{u+1} $$ which integrates into $\ln(u) - \ln(u+1)$ which is $\ln(e^x) - \ln(e^x+1)=x-\ln(e^x + 1)$


0

I am skeptical about a closed form for this definite integral. Its numerical value is $$0.20852064462921058579012825612057983198834121981775$$ which has not been identified by any inverse symbolic calculator. The closest found is $$\frac 1 {10}\Gamma \left(\frac{1}{6}\right) \Big(\log(2)\Big) ^{\Gamma \left(\frac{1}{3}\right)}=0.20852064115$$ What I suppose ...


4

To integrate $\displaystyle \int x^3 \sqrt{1 - x^2} dx$, I would most naturally use $u$-substitution with $u = 1 - x^2$ and $du = -2x\hspace{1mm}dx$. Then this is $$\begin{align} \int x^3 \sqrt{1 - x^2} dx &= \frac{-1}{2}\int (-2x\hspace{1mm}dx) (x^2) \sqrt{1 - x^2} \\ &= \frac{-1}{2}\int (1 - u) \sqrt{u} \; du \\ &= \frac{-1}{2} \bigg( \frac{u^{...


6

Using a u substition is very useful in this case $$\int x^3\sqrt{1-x^2}dx$$ $$u=1-x^2$$ $$du=-2x\hspace{1mm} dx$$ $$dx=-\frac{1}{2x} du$$ $$\int x^3\sqrt{u} \cdot -\frac{1}{2x}du$$ $$-\frac{1}{2} \int (1-u)\sqrt{u} \hspace{1mm}du$$ $$-\frac{1}{2} \int u^{\frac{1}{2}}-u^{\frac{3}{2}} \hspace{1mm}du$$ $$-\frac{1}{2} (\frac{2}{3}u^{\frac{3}{2}}-\frac{2}{...


9

Because the power of $x$ outside the radical is odd, this is most easily accomplished with the substitution $u = 1-x^2$ rather than a trig substitution. Then $du = -2x\,dx$ and \begin{multline} \int x^3\sqrt{1-x^2}dx = -\frac{1}{2}\int(1-u)\sqrt{u}\,du = -\frac{1}{2}\int\left(u^{1/2}-u^{3/2}\right)\,du = \frac{u^{5/2}}{5} - \frac{u^{3/2}}{3}\\ = \frac{(1-x^...


5

Using addition formulas you have: $\cosh(2y)-\cos(2x)=\cos(2iy)-\cos(2x)=2\sin(x+iy)\sin(x-iy)=2\sin(z)\sin(\bar z)$ Applying same kind of transformation to a numerator that would be symmetric, gives: $\sin(2x)-i\sinh(2y)=\sin(2x)-\sin(2iy)=2\cos(x+iy)\sin(x-iy)=2\cos(z)\sin(\bar z)$ Thus $f(z)=\dfrac 1{\tan(z)}$ is a candidate and since it is analytic, ...


1

First of all, your $g^{(-n)}$ is defined only up to a polynomial of degree $n-1$, so it's not quite correct to talk about the $n$-th integral. But, to make a definite choice, one may take definite integrals, say $$g^{(0)}(x):=g(x),\quad g^{(-n-1)}(x):=\int_{0}^{x}g^{(-n)}(y)\,dy,$$ and see what happens. We have in this case $\color{gray}{\text{[induction on $...


1

We cannot have $-1\le x\le1$ Let $\displaystyle I=\int \sqrt{\frac{x-1}{x+1}}\,\mathrm dx ,J=\int \sqrt{\frac{x+1}{x-1}}\,\mathrm dx$ $\displaystyle I+J=\int\frac{|x-1|+|x+1|}{\sqrt{x^2-1}}\,\mathrm dx, I-J=\int\frac{|x-1|-|x+1|}{\sqrt{x^2-1}}\,\mathrm dx$ Case $\#1:$ If $\displaystyle x>1,I+J=\int\frac{x-1+x+1}{\sqrt{x^2-1}}\,\mathrm dx, I-J=\int\...


1

As for calculus, $x\ge1$ or $x<-1$ WLOG $x=\sec2u,0\le2u<\pi$ $\tan2u=\pm\sqrt{x^2-1}$ The plus sign needs to be considered if $\tan2u\ge0$ if $0\le2u\le\dfrac\pi2$ $$I=\int\sqrt{\dfrac{x-1}{x+1}}dx=2\int\tan u\sec2u\tan2u\ du$$ $$=\int\dfrac{4\sin^2u}{\cos^22u}du$$ As $\cos2t=1-2\sin^2t,$ $$I=2\int(\sec^22u-\sec2u)du$$ Use How do I integrate $...


1

You can't. Check by yourself with the following counterexample Let $f(t)=t; g(t)=3.1$ and $h(x)=5$


5

No, you can't. Although, you can claim that $$ \int_{0}^{h(x)} e^{f(t)}(g(t) - f(t)) dt > 0, $$ which is indeed not the same as $$ \int_{0}^{h(x)} g(t) - f(t) dt > 0. $$ You can think of $e^{f(t)}$ as a weight-function.


1

$u = x$ is never a good substitution since if administered correctly, gives you exactly the same integral you started with---only with a different variable. In your case, you made a mistake by simply dropping the $x$ in your original problem which you must not do. Try integration by parts.


1

we have: $$I=\int x^m\ln(a+x)dx$$ now with $u=\ln(a+x)$ we get: $$I=\int(e^u-a)^me^uu\,du$$ now using integration by parts: $$I=\frac{u(e^u-a)^{m+1}}{m+1}-\int\frac{(e^u-a)^{m+1}}{m+1}du$$ now try using binomial expansion. One way of doing it would be by writing: $$(e^u-a)^{m+1}=e^{(m+1)u}(1-ae^{-u})^{m+1}$$


1

With one step of by-parts integration (on $x^m$), you get rid of the logarithm and reduce to an incomplete Beta integral. https://en.wikipedia.org/wiki/Beta_function (check the fifth property and the incomplete function). This indirectly proves that for general $m$ there is no closed-form expression.


3

Use integration by parts and note that $$\frac{x^{m+1}}{a+x}=\sum_{k=0}^m(-a)^kx^{m-k}+\frac{(-a)^{m+1}}{a+x}$$


2

HINT: $$ \begin{align} &\hphantom{=.}\int \ln(e^x\sin^3x)dx \\ &= \int \ln(e^x)+\ln(\sin^3x)dx \\ &= \int (x+3\ln(\sin x))dx \\ &= \frac{x^2}{2}+3\int \ln(\sin x)dx \end{align} $$


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