Get the solution to "Evaluate the integrals. ∫tan−1⁡xx2dx"

djeljenike

Answered question

2020-11-01

Evaluate the integrals.

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

pierretteA

Skilled2020-11-02Added 102 answers

Given

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

Apply integration by parts: \int f \cdot g^{'} = f g - \int f^{'} \cdot g

f = \tan^{- 1} (x), g^{'} = \frac{1}{x^{2}}

f^{'} = \frac{1}{x^{2} + 1}, g = - \frac{1}{x}

\int f \cdot g^{'} = - \frac{\tan^{- 1} (x)}{x} - \int - \frac{1}{x (x^{2} + 1)} d x

\int - \frac{1}{x (x^{2} + 1)} d x = - \int \frac{1}{x (x^{2} + 1)} d x

The integral solce by expanding the fraction by \frac{1}{x^{3}}

\int \frac{1}{x (x^{2} + 1)} d x = \int \frac{1}{(\frac{1}{x^{2}} + 1) x^{3}} d x

Substitute u = \frac{1}{x^{2}} + 1 \to \frac{d u}{d x} = - \frac{2}{x^{3}} \to d x = - \frac{x^{3}}{2} d u

\int \frac{1}{(\frac{1}{x^{2}} + 1) x^{3}} d x = \int \frac{1}{(u) x^{3}} \times - \frac{x^{3}}{2} d u = - \frac{1}{2} \int \frac{1}{u} d u

= - \frac{\ln (u)}{2}

Undo substitution: u = \frac{1}{x^{2}} + 1

= - \frac{\ln (\frac{1}{x^{2}} + 1)}{2}

\Rightarrow - \int \frac{1}{x (x^{2} + 1)} d x = \frac{\ln (\frac{1}{x^{2}} + 1)}{2}

\Rightarrow - \frac{\tan^{- 1} (x)}{x} - \int - \frac{1}{x (x^{2} + 1)} d x = \frac{\tan^{- 1} (x)}{x} - \frac{\ln (\frac{1}{x^{2}} + 1)}{2} + C

\int \frac{\tan^{- 1} x}{x^{2}} d x = \frac{\tan^{- 1} (x)}{x} - \frac{\ln (\frac{1}{x^{2}} + 1)}{2} + C

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?