1. ## Improper Integral

Hi I need help on an improper integral.

The integral from negative infinity to 3, where f(x)= 1/(x^2+9)

I know I need to get it to the format, x^2+1 on the bottom, in order to get it to a tangent inverse; however, I am having trouble with my algebra to get to this point. Moreover we cannot use u-sub, because y' of x^2+9= 2x, and we cannot make an X; so this is our only option.

2. ## Re: Improper Integral

Divide both numerator and denominator by 9: $\displaystyle \frac{1}{x^2+ 9}= \frac{\frac{1}{9}}{\left(\frac{x}{3}\right)^2+1}$ so that the substitution $\displaystyle y= \frac{x}{3}$, $\displaystyle dy= \frac{1}{3}dx$, and then $\displaystyle 3dy= dx$, gives $\displaystyle \int_{-\infty}^3\frac{dx}{x^3+ 9}= 3\int_{-\infty}^1 \frac{dy}{y^2+ 1}$.

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