fanyattehedzg

2021-12-26

Evaluate the integrals.
$\int \frac{{\left({\mathrm{tan}}^{-1}x\right)}^{5}}{\left(1+{x}^{2}\right)}dx$

Lakisha Archer

Step 1
We have to evaluate the integral:
$\int \frac{{\left({\mathrm{tan}}^{-1}x\right)}^{5}}{\left(1+{x}^{2}\right)}dx$
This integral will be solved by substitution method since derivative of one function is present in the integral.
Assuming,
$t={\mathrm{tan}}^{-1}x$
Differentiating with respect to x, we get
$\frac{dt}{dx}=\frac{d\left({\mathrm{tan}}^{-1}x\right)}{dx}$
$=\frac{1}{1+{x}^{2}}$
$dt=\frac{1}{1+{x}^{2}}dx$
Step 2
Substituting above values in the integral,
$\int \frac{{\left({\mathrm{tan}}^{-1}x\right)}^{5}}{\left(1+{x}^{2}\right)}dx=\int {t}^{5}dt$
$=\frac{{t}^{5+1}}{5+1}+C$ (since $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$)
$=\frac{{t}^{6}}{6}+C$
$=\frac{1}{6}{\left({\mathrm{tan}}^{-1}x\right)}^{6}+C$
Where, C is an arbitrary constant.
We have substituted value of t for last step.
Hence, value of integral is $\frac{1}{6}{\left({\mathrm{tan}}^{-1}x\right)}^{6}+C$.

David Clayton

$\int \frac{{\left({\mathrm{tan}}^{-1}x\right)}^{5}}{\left(1+{x}^{2}\right)}dx=\int \frac{{\mathrm{arctan}}^{5}\left(x\right)}{{x}^{2}+1}dx$
$\int \frac{{\mathrm{arctan}}^{5}\left(x\right)}{{x}^{2}+1}dx$
Substitution $u=\mathrm{arctan}\left(x\right)⇒\frac{du}{dx}=\frac{1}{{x}^{2}+1}⇒dx=\left({x}^{2}+1\right)du:$
$=\int {u}^{5}du$
Integral of a power function:
$\int {u}^{n}du=\frac{{u}^{n+1}}{n+1}$ at n=5:
$=\frac{{u}^{6}}{6}$
Reverse replacement $u=\mathrm{arctan}\left(x\right):$
$=\frac{{\mathrm{arctan}}^{6}\left(x\right)}{6}$
$\int \frac{{\mathrm{arctan}}^{5}\left(x\right)}{{x}^{2}+1}dx$
$=\frac{{\mathrm{arctan}}^{6}\left(x\right)}{6}+C$

karton

$\int \frac{\left({\mathrm{tan}}^{-1}x{\right)}^{5}}{\left(1+{x}^{2}\right)}dx=\int \frac{{\mathrm{arctan}}^{5}\left(x\right)}{{x}^{2}+1}dx$
Transform the expression
$\int {t}^{5}dt$
Use $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1},n\ne -1$ to evaluate the integral
$\frac{{t}^{6}}{6}$
Substitute back
$\frac{\mathrm{arctan}\left(x{\right)}^{6}}{6}$
$\frac{\mathrm{arctan}\left(x{\right)}^{6}}{6}+C$