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

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

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

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

2 ln x^{2} + 1 - 3 arctan (x) + C

Solution steps

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

Expand

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

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

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

\int \frac{3}{x ^{2} + 1} d x = 3 arctan (x)

= 2 ln x^{2} + 1 - 3 arctan (x)

Add a constant to the solution

= 2 ln x^{2} + 1 - 3 arctan (x) + C

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