a) Intergrate (4x+5)/(x^2 +2x + 2)

b) Intergrate x^2 inversetan x

thanks =)

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- Aug 11th 2009, 01:04 AMduskmantleIntergration problem
a) Intergrate (4x+5)/(x^2 +2x + 2)

b) Intergrate x^2 inversetan x

thanks =) - Aug 11th 2009, 01:19 AMFailure
Ad a) consider $\displaystyle \frac{4x+5}{x^2+2x+2}=2\cdot \frac{2x+2}{x^2+2x+2}+\frac{1}{(x+1)^2+1}$. Both terms on the left hand side are easy to integrate (by substitution).

Ad b) I suggest using partial integration (integrate $\displaystyle x^2$, differentiate $\displaystyle \arctan(x)$), then do a polynomial division, finally apply a substition $\displaystyle u := 1+x^2$. - Aug 11th 2009, 01:42 AMduskmantle
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, 01:54 AMFailure
Integrating $\displaystyle -\frac{1}{6x}$ would be easy

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

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

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

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