What is the integral of ((x^2+13))/(x^2+16) ?

The integral of ((x^2+13))/(x^2+16) is -3/4 arctan(x/4)+x+C

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integral of ((x^2+13))/(x^2+16)

\int \frac{( x ^{2} + 13 )}{x ^{2} + 16} d x

- \frac{3}{4} arctan (\frac{x}{4}) + x + C

Solution steps

\int \frac{x ^{2} + 13}{x ^{2} + 16} d x

\frac{x ^{2} + 13}{x ^{2} + 16} = - \frac{3}{x ^{2} + 16} + 1

= \int - \frac{3}{x ^{2} + 16} + 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} + 16} d x + \int 1 d x

\int \frac{3}{x ^{2} + 16} d x = \frac{3}{4} arctan (\frac{x}{4})

\int 1 d x = x

= - \frac{3}{4} arctan (\frac{x}{4}) + x

Add a constant to the solution

= - \frac{3}{4} arctan (\frac{x}{4}) + x + C

What is the integral of ((x^2+13))/(x^2+16) ?
The integral of ((x^2+13))/(x^2+16) is -3/4 arctan(x/4)+x+C