Solved: "Evaluate the following integrals: ∫x2tan−1xdx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

rescuedbyhimw0

Answered question

2021-12-03

Evaluate the following integrals:

\int x^{2} \tan^{- 1} x d x

Answer & Explanation

Witheyesse47

Beginner2021-12-04Added 14 answers

Step 1
Consider the integral: $\int x^{2} \tan^{- 1} x d x$ .
Apply Integration by parts $\int u v^{'} = u v - \int u^{'} v$ . Let $u = \tan^{- 1} x$ and $v^{'} = x^{2}$ .
Implies $u^{'} = \frac{1}{1 + x^{2}}$ and $v = \frac{x^{3}}{3}$ .
Solve Integral
$\int u^{'} v = \frac{1}{3} \int \frac{x^{3}}{x^{2} + 1} d x$
Let $x^{2} + 1 = u$ implies $2 x d x = d u$ . Substitute the values
$\frac{1}{3} \int \frac{x^{3}}{x^{2} + 1} d x = \frac{1}{3} \int \frac{u - 1}{2 u} d u$
$= \frac{1}{6} (u - \ln u)$
$= \frac{1}{6} (x^{2} + 1 - \ln (x^{2} + 1))$
Step 2
Substitute u,v and $\int$ u'v in the byparts formula.
$\int x^{2} \tan^{- 1} x d x = \frac{x^{3} \tan^{- 1} x}{3} - \frac{(x^{2} + 1)}{6} + \frac{\ln (x^{2} + 1)}{6} + C$
$= \frac{2 x^{3} \tan^{- 1} x - (x^{2} + 1) + \ln (x^{2} + 1)}{6} + C$

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