# Trig substitution integration?

## Homework Statement

Integrate dx/((x^2+1)^2)

Tan^2=sec^2-1

## The Attempt at a Solution

So I let x=tanx then dx=sec^2x

Then plugging everything in;

Sec^2(x)/(tan^2+1)^2

So it's sec^2/(sec^2x)^2) which is sec^2x/sec^4x

Canceling out the sec^2 gives you

1/sec^2x

Integrate: ln(sec^2x)+C

Is that right?

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LCKurtz
Homework Helper
Gold Member

## Homework Statement

Integrate dx/((x^2+1)^2)

Tan^2=sec^2-1

## The Attempt at a Solution

So I let x=tanx then dx=sec^2x
Don't use the same letter in a change of variables. Let ##y=\tan x## so ##dy=\sec^2x dx##.
[Edit, corrections follow] I meant ##x = \tan y## so ##dx =\sec^2 y dy##.
Then plugging everything in;

Sec^2(x)/(tan^2+1)^2

So it's sec^2/(sec^2x)^2) which is sec^2x/sec^4x

Canceling out the sec^2 gives you

1/sec^2x

Integrate: ln(sec^2x)+C

Is that right?
No. That isn't its antiderivative.

Last edited:
benorin
Homework Helper
Gold Member

## Homework Statement

Integrate dx/((x^2+1)^2)

Tan^2=sec^2-1

## The Attempt at a Solution

So I let x=tanx then dx=sec^2x

Then plugging everything in;

Sec^2(x)/(tan^2+1)^2

So it's sec^2/(sec^2x)^2) which is sec^2x/sec^4x

Canceling out the sec^2 gives you

1/sec^2x

Integrate: ln(sec^2x)+C

Is that right?
Like the man said, use another variable: Let $x=\tan\theta\Rightarrow dx=\sec^2\theta d\theta$ and the integral becomes

$$\int\frac{dx}{\left(1+x^2\right) ^2}=\int\frac{\sec^2\theta d\theta}{\left(1+\tan^2 x\right) ^2}=\int\frac{d\theta}{\sec^2\theta}=\int\cos^2\theta\, d\theta$$

now use the identity $\cos^2\theta=\frac{1}{2}\left( 1+\cos 2\theta\right)$ to integrate w.r.t. $\theta$ then use $x=\tan\theta$ to rewrite the functions of $\theta$ as functions of $x$.