What is the integral of (9+4x)/(1+x^2) ?

The integral of (9+4x)/(1+x^2) is 9arctan(x)+2ln|1+x^2|+C

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integral of (9+4x)/(1+x^2)

\int \frac{9 + 4 x}{1 + x ^{2}} d x

9 arctan (x) + 2 ln 1 + x^{2} + C

Solution steps

\int \frac{9 + 4 x}{1 + x ^{2}} d x

Expand

\frac{9 + 4 x}{1 + x ^{2}} : \frac{9}{1 + x ^{2}} + \frac{4 x}{1 + x ^{2}}

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{9}{1 + x ^{2}} d x + \int \frac{4 x}{1 + x ^{2}} d x

\int \frac{9}{1 + x ^{2}} d x = 9 arctan (x)

\int \frac{4 x}{1 + x ^{2}} d x = 2 ln 1 + x^{2}

= 9 arctan (x) + 2 ln 1 + x^{2}

Add a constant to the solution

= 9 arctan (x) + 2 ln 1 + x^{2} + C

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What is the integral of (9+4x)/(1+x^2) ?
The integral of (9+4x)/(1+x^2) is 9arctan(x)+2ln|1+x^2|+C