# Intergration problem

• Aug 11th 2009, 02:04 AM
duskmantle
Intergration problem
a) Intergrate (4x+5)/(x^2 +2x + 2)

b) Intergrate x^2 inversetan x

thanks =)
• Aug 11th 2009, 02:19 AM
Failure
Quote:

Originally Posted by duskmantle
a) Intergrate (4x+5)/(x^2 +2x + 2)

b) Intergrate x^2 inversetan x

thanks =)

Ad a) consider $\frac{4x+5}{x^2+2x+2}=2\cdot \frac{2x+2}{x^2+2x+2}+\frac{1}{(x+1)^2+1}$. Both terms on the left hand side are easy to integrate (by substitution).

Ad b) I suggest using partial integration (integrate $x^2$, differentiate $\arctan(x)$), then do a polynomial division, finally apply a substition $u := 1+x^2$.
• Aug 11th 2009, 02:42 AM
duskmantle
um i got part a thanks

but part b i ended up having to intergrate -1/6x ...how do i intergrate that? =(
• Aug 11th 2009, 02:54 AM
Failure
Quote:

Originally Posted by duskmantle
um i got part a thanks

but part b i ended up having to intergrate -1/6x ...how do i intergrate that? =(

Integrating $-\frac{1}{6x}$ would be easy

$\int -\frac{1}{6x}\,dx =-\frac{1}{6}\int \frac{1}{x}\,dx=-\frac{1}{6}\cdot\ln|x|+C$

However, I think you don't need that integral here, becaues I get

$\int \underset{\uparrow}{x^2}\cdot \underset{\downarrow}{\tan^{-1}(x)}\, dx=\frac{1}{3}x^3\tan^{-1}(x)-\frac{1}{3}\cdot\int x^3\frac{1}{1+x^2}\, dx$

And now you have $\frac{x^3}{1+x^2}=x-\frac{x}{1+x^2}$. Integrating the first term is trivial and integrating the second term is done by substitution $u:=1+x^2$.