Integral with trig substitution

[hahaha158]

Homework Statement

∫(x+1)/((x^2+1)^2)

Homework Equations

The Attempt at a Solution

I have been able to separate this into 2

∫x/(x^2+1)^2 dx which i found to be equal to (1/2)arctanx

and

∫1/(x^2+1)^2 dx which i am unable to find

What i did was sub in x=tanθ and dx=sec^2(θ)dθ, and with this i was able to get the integral to the form

∫1/sec^2(θ)dθ=∫cos^2(θ)dθ =

∫(1+cos2θ)/2

I then separated this into two integrals

∫1/2=θ/2

and

∫cos2θ/2 = sin2θ/4

However, the answer given is (x-1)/(2(x^2+1)+C

I am not sure how to get to this answer from

θ/2+sin2θ/4+C, i tried subbing in θ=arctanx and i get a very messy equation

arctanx/2+sin(2arctanx)/4+C

I was able to simplify it to

cosx/2sinx+sin(2cosx/sinx)/4+C, and i can't get any farther twoards the answer.

Can anyone please explain how i get there or did i make a mistake somewhere in my work?

Thanks very much

 

[LCKurtz]

For what it's worth, their answer is certainly wrong because it surely must have an arctan(x) term. Maple gives:

[(1/4)(4x-2)/(x^2+1)] + arctan(x)

 

[hahaha158]

For what it's worth, their answer is certainly wrong because it surely must have an arctan(x) term. Maple gives:

[(1/4)(4x-2)/(x^2+1)] + arctan(x)

oops i entered the original question incorrectly, it should be 1+x on the top rather than 2+x, also the answer i wrote was for one part of the integral only, not the ansewr to the whole question, i should have clariified sorry.

 

[SammyS]

Homework Statement

∫(x+1)/((x^2+1)^2)

Homework Equations

The Attempt at a Solution

I have been able to separate this into 2

∫x/(x^2+1)^2 dx which i found to be equal to (1/2)arctanx

and

∫1/(x^2+1)^2 dx which i am unable to find
...

Thanks very much

\displaystyle \int \frac{x}{(x^2+1)^2}\,dx\ne\frac{1}{2}\arctan(x)

Use substitution to evaluate this integral. Let u = x² + 1 ...


For \displaystyle \int \frac{1}{(x^2+1)^2}\,dx\,,\ write the integrand as \displaystyle \frac{x^2+1}{(x^2+1)^2}-\frac{x^2}{(x^2+1)^2}\ .\

The first term gives the arctan part of the result. For the second term, do integration by parts, with u = x, dv = ...