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Math Help - Check my work with these integrals...

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

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

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

    \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 \int\frac{1}{a^2+u^2}du = \frac{1}{a}\tan^{-1}\frac{u}{a} + c

    So, I was left with:

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

    Solving the remaining integral:

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

    And substitute back:

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

    Thoughts?

    Thank you!
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  2. #2
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    Quote Originally Posted by Polyxendi View Post
    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. \int \frac{x+a}{x^2 + a^2}dx

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

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

    \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 \int\frac{1}{a^2+u^2}du = \frac{1}{a}\tan^{-1}\frac{u}{a} + c

    So, I was left with:

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

    Solving the remaining integral:

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

    And substitute back:

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

    Thoughts?

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

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