Don't know how to work out the following integral.
Substituting cos x^5 = t does not seem to work.
This looks much harder than it is.
Let $\displaystyle u=x^5, du=5x^4dx$
Another way of solving for "du" is $\displaystyle \frac{1}{5}du=x^4dx$. But we don't want just x^4dx, we want 2x^4, so multiply by 2 on both sides to get $\displaystyle \frac{2}{5}du=2x^4$. Now we're getting somewhere.
So plug everything back in. Our new integral in terms of u is...
$\displaystyle \frac{2}{5} \int \frac{1}{\cos^2(u)}du$
Now what's $\displaystyle \frac{1}{\cos^2(x)}$?? $\displaystyle \sec^2(x)$.
See it now?
Hello,
your problem looks harder than it is actually:
$\displaystyle \int{\frac{dx}{x\cdot \sqrt[5]{(\ln(x)+2)^2}}}=\int{\frac{1}{x} \cdot (\ln(x)+2)^{-\frac{2}{5}}dx}$
let $\displaystyle u=\ln(x)+2$. Then
$\displaystyle \frac{du}{dx}=\frac{1}{x}$. Thus $\displaystyle du=\frac{1}{x} dx$.
Now plug in these terms into the original integral:
$\displaystyle \int{\frac{dx}{x\cdot \sqrt[5]{(\ln(x)+2)^2}}}=\int{u^{-\frac{2}{5}} du}$
I'll leave the rest for you.
(For confirmation only: $\displaystyle \frac{5}{3}(\ln(x)+2)^{\frac{3}{5}}$
EB
One thing I hate about integral is checking differenciability intervals and domains. Since I am lazy here was my approach.
$\displaystyle \int \tan^{-1} x dx$
Express as,
$\displaystyle \int (1+x^2) \cdot \frac{\tan^{-1} x}{1+x^2} dx$
Now,
$\displaystyle 1+x^2=\sec ^2 (\tan^{-1} x)$
Thus,
$\displaystyle \int \sec^2 (\tan^{-1} x) \cdot \tan^{-1} x\cdot \frac{1}{1+x^2} dx$
Let,
$\displaystyle f(x)=x\sec^2x$
$\displaystyle g(x)=\tan^{-1} x$
$\displaystyle g'(x)=1/(1+x^2)$
Thus,
$\displaystyle \int (f \circ g)\cdot g' dx=F\circ g+C$
Where,
$\displaystyle F$ is an anti-derivative of $\displaystyle x\sec^2 x$
So the problem reduces to finding,
$\displaystyle \int x\sec^2 xdx$
Which should be doable by integration by parts.
$\displaystyle \int \frac{xdx}{1+x^2}=\frac{1}{2}\int\frac{d(1+x^2)}{1 +x^2}=\frac{1}{2}ln|1+x^2|+C$
Where does $\displaystyle \frac{1}{2}$ come from ?
Where does $\displaystyle d(1+x^2)$ come from ?
If $\displaystyle \int\frac{dx}{1+x^2}=arctanx+C$ how is it possible that my answer has $\displaystyle ln$ ?
I would start off with integration by parts:
$\displaystyle
\int {\arctan x} dx = x\arctan x - \int {\frac{x}{{1 + x^2 }}dx}
$
Now, the second integral is easy:
$\displaystyle
\int {\frac{x}{{1 + x^2 }}dx} = \frac{1}{2}\int {\frac{1}{{1 + x^2 }}d\left( {1 + x^2 } \right) = \frac{1}{2}\ln \left| {1 + x^2 } \right|} + C
$
So combined:
$\displaystyle
\int {\arctan x} dx = x\arctan x - \frac{1}{2}\ln \left| {1 + x^2 } \right| + C
$