Question

hi, i'm integrating (2x^2 +3)/(x^2 +1)^2
and i'd reach the following step and get stucked =='
3/2tan^-1 (x) + 1/4 sin [2(tan^-1 (x))]

Answer 1

basically its the sin of 2u, the u i previously let to be tan (x), now it became sin of 2 tan^-1 x

Answer 2

my calculus foundation sux to the max ==

Answer 3

What I would do is break it up:
\[2 \int\limits \frac{x^2 dx}{(x^2+1)^2}+3 \int\limits \frac{dx}{(x^2+1)^2}\]

From here you can do trig sub:
Let
\[\tan(\Omega)=x; \sec^2(\Omega) d \Omega= dx; \sec(\Omega)=\sqrt{x^2+1}\]
All of that is coming from the pythagorean theorem.
Replace this in the integral
\[\sec(\Omega)=\sqrt{x^2+1} \implies \sec^4(\Omega)=(x^2+1)^2\]
\[2 \int\limits \frac{\tan^2(\Omega)\sec^2(\Omega)}{\sec^4(\Omega)}d \Omega+3 \int\limits \frac{\sec^2(\Omega)}{\sec^4(\Omega)} d \Omega\]
\[2 \int\limits \frac{\sin^2(\Omega)}{\cos^2(\Omega)}\cos^2(\Omega) d \Omega + 3 \int\limits \cos^2(\Omega) d \Omega\]
Do you follow any of this?

Answer 4

I guess partial fractions would have been easier >.>

Answer 5

yea i approached it with partial fractions, the integral of the 1st term with x^2 +1 i got 2 tan^-1 x, but i think im stucked at the (x^2+1)^2 part

Answer 6

erm, how the 3 is formed at the 2nd partial fraction term? i got 1 on top o.o

Answer 7

Okay, the do partial fractions:
\[\frac{2x^2+3}{(x^2+1)^2}=\frac{Ax+\eta}{x^2+1}+\frac{\xi x + B}{(x^2+1)^2}\]
Getting a common denominator implies that
\[A=0; B=0; \eta = 2; \xi = 1\]
Gives:
\[I=2 \int\limits \frac{dx}{x^2+1}+ \int \frac{dx}{(x^2+1)^2}\]
The first term is indeed 2 arctan:
\[I=2 \arctan(x)+\int\limits \frac{dx}{(x^2+1)^2}\]

Answer 8

Is this where you are stuck?

Answer 9

not sure whether the 2nd part i integrated wrongly or not, cant merge all the 3 terms
2nd part i get \[1/4 \sin 2\left( \tan^{-1} x \right) + 1/2 \tan^{-1} \]

Answer 10

or its just me, dunoe how to simplify them ==

Answer 11

Also:
tan(omega)=opp/adj=x
So:
\[\sec^2(\Omega) \frac{d \Omega}{dx}=1 \implies \sec^2(\Omega) d \Omega= dx\]
Also
sec(omega)=1/cos(omega)=1/(adj/hyp)=hyp/adj=sqrt{1+x^2}/1
So we have:
\[\tan(\xi)=x; \sec^2(\xi) d \xi = dx; \sec(\xi)=\sqrt{1+x^2}\]