What is the integral of (x^2+61)/(x^2+64) ?

The integral of (x^2+61)/(x^2+64) is -3/8 arctan(x/8)+x+C

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integral of (x^2+61)/(x^2+64)

\int \frac{x ^{2} + 61}{x ^{2} + 64} d x

- \frac{3}{8} arctan (\frac{x}{8}) + x + C

Solution steps

\int \frac{x ^{2} + 61}{x ^{2} + 64} d x

\frac{x ^{2} + 61}{x ^{2} + 64} = - \frac{3}{x ^{2} + 64} + 1

= \int - \frac{3}{x ^{2} + 64} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{3}{x ^{2} + 64} d x + \int 1 d x

\int \frac{3}{x ^{2} + 64} d x = \frac{3}{8} arctan (\frac{x}{8})

\int 1 d x = x

= - \frac{3}{8} arctan (\frac{x}{8}) + x

Add a constant to the solution

= - \frac{3}{8} arctan (\frac{x}{8}) + x + C

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What is the integral of (x^2+61)/(x^2+64) ?
The integral of (x^2+61)/(x^2+64) is -3/8 arctan(x/8)+x+C