What is the integral of (3+6x)/(1+x^2) ?

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

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integral of (3+6x)/(1+x^2)

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (x \cdot ln (x + 1))

\frac{d}{d x} (\frac{4}{1 + x})

\int \frac{8}{x ( ln ( x ) + 1 ) ^{\frac{3}{2}}} d x

3 x - 4

\int sin (t - s) d s

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