What is the integral of (3x+4)/(1+x^2) ?

The integral of (3x+4)/(1+x^2) is 3/2 ln|1+x^2|+4arctan(x)+C

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integral of (3x+4)/(1+x^2)

\int \frac{3 x + 4}{1 + x ^{2}} d x

\frac{3}{2} ln 1 + x^{2} + 4 arctan (x) + C

Solution steps

\int \frac{3 x + 4}{1 + x ^{2}} d x

Expand

\frac{3 x + 4}{1 + x ^{2}} : \frac{3 x}{1 + x ^{2}} + \frac{4}{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{3 x}{1 + x ^{2}} d x + \int \frac{4}{1 + x ^{2}} d x

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

\int \frac{4}{1 + x ^{2}} d x = 4 arctan (x)

= \frac{3}{2} ln 1 + x^{2} + 4 arctan (x)

Add a constant to the solution

= \frac{3}{2} ln 1 + x^{2} + 4 arctan (x) + C

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What is the integral of (3x+4)/(1+x^2) ?
The integral of (3x+4)/(1+x^2) is 3/2 ln|1+x^2|+4arctan(x)+C