# Thread: Check my work with these integrals...

1. ## Check my work with these integrals...

I had a Calc 2 test today and I wanted to make sure that I was on the right track. There are two, specifically, that worry me. Here is the first with what I did. I'll post the other later.

1. $\displaystyle \int \frac{x+a}{x^2 + a^2}dx$

$\displaystyle \int\frac{x}{x^2 + a^2}dx + \int\frac{a}{x^2+a^2}dx$

I used $\displaystyle u = x^2 + a^2$ and $\displaystyle du = 2xdx$ for the first expression and was left with:

$\displaystyle \frac{1}{2}\int\frac{du}{u} + a\int\frac{1}{x^2+a^2}dx$

Then, on the second expression I used the inverse trigonometric identity of $\displaystyle \int\frac{1}{a^2+u^2}du = \frac{1}{a}\tan^{-1}\frac{u}{a} + c$

So, I was left with:

$\displaystyle \frac{1}{2}\int\frac{du}{u} + \frac{1}{a}\tan^{-1}\frac{x}{a} + c$

Solving the remaining integral:

$\displaystyle \frac{1}{2}\ln|u| + \frac{1}{a}\tan^{-1}\frac{x}{a} + c$

And substitute back:

$\displaystyle \frac{1}{2}\ln|x^2+a^2| + \frac{1}{a}\tan^{-1}\frac{x}{a} + c$

Thoughts?

Thank you!

2. Originally Posted by Polyxendi
I had a Calc 2 test today and I wanted to make sure that I was on the right track. There are two, specifically, that worry me. Here is the first with what I did. I'll post the other later.

1. $\displaystyle \int \frac{x+a}{x^2 + a^2}dx$

$\displaystyle \int\frac{x}{x^2 + a^2}dx + \int\frac{a}{x^2+a^2}dx$

I used $\displaystyle u = x^2 + a^2$ and $\displaystyle du = 2xdx$ for the first expression and was left with:

$\displaystyle \frac{1}{2}\int\frac{du}{u} + a\int\frac{1}{x^2+a^2}dx$

Then, on the second expression I used the inverse trigonometric identity of $\displaystyle \int\frac{1}{a^2+u^2}du = \frac{1}{a}\tan^{-1}\frac{u}{a} + c$

So, I was left with:

$\displaystyle \frac{1}{2}\int\frac{du}{u} + \frac{1}{a}\tan^{-1}\frac{x}{a} + c$

Solving the remaining integral:

$\displaystyle \frac{1}{2}\ln|u| + \frac{1}{a}\tan^{-1}\frac{x}{a} + c$

And substitute back:

$\displaystyle \frac{1}{2}\ln|x^2+a^2| + \frac{1}{a}\tan^{-1}\frac{x}{a} + c$

Thoughts?

Thank you!
you forgot the factor $\displaystyle a$ in front of the integral that yields the inverse tangent function ...

$\displaystyle \ln{\sqrt{x^2+a^2}} + \arctan\left(\dfrac{x}{a}\right) + C$