What is the integral of (4x+6)/(x^2+1) ?

The integral of (4x+6)/(x^2+1) is 2ln|x^2+1|+6arctan(x)+C

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integral of (4x+6)/(x^2+1)

\int \frac{4 x + 6}{x ^{2} + 1} d x

2 ln x^{2} + 1 + 6 arctan (x) + C

Solution steps

\int \frac{4 x + 6}{x ^{2} + 1} d x

Expand

\frac{4 x + 6}{x ^{2} + 1} : \frac{4 x}{x ^{2} + 1} + \frac{6}{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{6}{x ^{2} + 1} d x

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

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x)

= 2 ln x^{2} + 1 + 6 arctan (x)

Add a constant to the solution

= 2 ln x^{2} + 1 + 6 arctan (x) + C

(4 x^{2}) (e^{- 4 x})

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

\int (x^{2} - 9 x + 5) d x

x \to + 1 lim (x^{2})

\frac{- 6 x ^{2} - 4 x + 4}{( 3 x ^{2} + 2 ) ^{2}}

What is the integral of (4x+6)/(x^2+1) ?
The integral of (4x+6)/(x^2+1) is 2ln|x^2+1|+6arctan(x)+C