- #1

- 832

- 30

## Homework Statement

Evaluate the next integrals, expressing it previously in the forms that contains [tex]\sqrt[ ]{a^2+u^2}[/tex], [tex]\sqrt[ ]{a^2-u^2}[/tex] and [tex]\sqrt[ ]{u^2-a^2}[/tex], and then solve it using the integral table;

The exercise which I couldn't solve:

[tex]\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sqrt[ ]{x^2+2x}}dx[/tex]

## The Attempt at a Solution

I started by completing the square:

[tex]x^2+2x=(x+1)^2-1[/tex]

Then

[tex]u^2=(x+1)^2[/tex]

[tex]u=(x+1)\Rightarrow{x+3=u+2}[/tex]

[tex]du=dx[/tex]

[tex]\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sqrt[ ]{x^2+2x}}dx=\displaystyle\int_{}^{}\displaystyle\frac{u+2}{\sqrt[ ]{u^2-1}}du[/tex]

From here I've tried to solve it by parts

[tex]t=u+2[/tex]

[tex]dt=du[/tex]

[tex]dv=\displaystyle\frac{1}{\sqrt[ ]{u^2-1}}du[/tex]

[tex]v=\ln|u+\sqrt[ ]{u^2-1}|[/tex]

[tex]\displaystyle\int_{}^{}\displaystyle\frac{u+2}{\sqrt[ ]{u^2-1}}du=(u+2)\ln|u+\sqrt[ ]{u^2-1}|-\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du[/tex]

If I try again by parts with [tex]-\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du[/tex] it gets more complicated.

So, what do you say?

Bye there.