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

The integral of (6+6x)/(1+x^2) is 6arctan(x)+3ln|1+x^2|+C

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

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

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

Solution steps

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

Expand

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

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

\int \frac{6 x}{1 + x ^{2}} d x = 3 ln 1 + x^{2}

= 6 arctan (x) + 3 ln 1 + x^{2}

Add a constant to the solution

= 6 arctan (x) + 3 ln 1 + x^{2} + C

- \frac{13}{s ^{5}}

ln (1 + x)

\frac{d}{d x} (\frac{1}{a x + b})

\int - 2 sin^{3} (x) cos (x) d x

\int \frac{5 t - 2}{t + 1} d t

What is the integral of (6+6x)/(1+x^2) ?
The integral of (6+6x)/(1+x^2) is 6arctan(x)+3ln|1+x^2|+C