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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{4}{1 + x ^{2}} d x = 4 arctan (x)

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

= 4 arctan (x) + 2 ln 1 + x^{2}

Add a constant to the solution

= 4 arctan (x) + 2 ln 1 + x^{2} + C

f (x) = \frac{2 x + 1}{x ^{2} + 2}

(- 2.2) (3.4)

\int (1 + x) e^{- x} d x

\frac{d}{d x} (\frac{x ^{7}}{25})

8 x^{2} y^{^{'}} = y^{^{'}} + 7 x e^{- y}

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