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Ikunupe6v Answered2022-01-05

I am in the middle of a problem and having trouble integrating the following integral:

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

Answer & Explanation

puhnut1m

Expert

2022-01-06Added 33 answers

Try making a substitution

x = \tan u

. Notice then that

{(1 + x^{2})}^{2} = {(1 + \tan^{2} u)}^{2} = {(\sec^{2} u)}^{2} = {(\sec^{2} u)}^{2} = \sec^{4} u

and

d x = \sec^{2} u d u

So the indefinite integral is now

\int \frac{1}{\sec^{2} u} d u = \int \cos^{2} u d u

This new integrand should be easier to integrate. Just remember to change your limits to get the proper evaluation.

Robert Pina

Expert

2022-01-07Added 42 answers

If we add and subtract

x^{2}

in the numerator, we can integrate the first integral immediately

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

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

and the second integral by parts:

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

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

user_27qwe

Expert

2022-01-11Added 230 answers

$\frac{1}{(1 + x^{2})^{2}} = (\frac{1}{1 + x^{2}} - \frac{x^{2}}{1 + x^{2})^{2}}) = (\frac{1}{1 + x^{2}} + \frac{x}{2} (\frac{1}{1 + x^{2}})^{'}) = (\frac{1}{1 + x^{2}} + (\frac{1}{2} \frac{x}{1 + x^{2}}) - \frac{1}{2} \frac{1}{1 + x^{2}}) = \frac{1}{2} (\arctan x + \frac{x}{1 + x^{2}})^{'}$