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

The integral of 1/((1+2x)(1+x^2)) is 2/5 ln|2x+1|+1/5 (-ln|x^2+1|+arctan(x))+C

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

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

\frac{2}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} (- ln x^{2} + 1 + arctan (x)) + C

Solution steps

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

Take the partial fraction of

\frac{1}{( 1 + 2 x ) ( 1 + x ^{2} )} : \frac{4}{5 ( 2 x + 1 )} + \frac{- 2 x + 1}{5 ( x ^{2} + 1 )}

= \int \frac{4}{5 ( 2 x + 1 )} + \frac{- 2 x + 1}{5 ( x ^{2} + 1 )} d x

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}{5 ( 2 x + 1 )} d x + \int \frac{- 2 x + 1}{5 ( x ^{2} + 1 )} d x

\int \frac{4}{5 ( 2 x + 1 )} d x = \frac{2}{5} ln ∣ 2 x + 1 ∣

\int \frac{- 2 x + 1}{5 ( x ^{2} + 1 )} d x = \frac{1}{5} (- ln x^{2} + 1 + arctan (x))

= \frac{2}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} (- ln x^{2} + 1 + arctan (x))

Add a constant to the solution

= \frac{2}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} (- ln x^{2} + 1 + arctan (x)) + C

\int \frac{2 - x + 1}{2 + x + 1} d x

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

\int \frac{sin ^{3} ( 4 x )}{cos ( 4 x )} d x

\int \frac{sin ( x )}{( cos ( x ) ) ^{5}} d x

3 x^{2}

What is the integral of 1/((1+2x)(1+x^2)) ?
The integral of 1/((1+2x)(1+x^2)) is 2/5 ln|2x+1|+1/5 (-ln|x^2+1|+arctan(x))+C