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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

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

\int 2 1 - x d x

\int x^{3} + 3 x^{2} - 2 x d x

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

\int e^{2 x} - e^{- 2 x} d x

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