What is the integral of (6x^2+5x+4)/((1+x)(1+4x^2)) ?

The integral of (6x^2+5x+4)/((1+x)(1+4x^2)) is ln|x+1|+1/4 ln|4x^2+1|+3/2 arctan(2x)+C

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integral of (6x^2+5x+4)/((1+x)(1+4x^2))

\int \frac{6 x ^{2} + 5 x + 4}{( 1 + x ) ( 1 + 4 x ^{2} )} d x

ln ∣ x + 1 ∣ + \frac{1}{4} ln 4 x^{2} + 1 + \frac{3}{2} arctan (2 x) + C

Solution steps

\int \frac{6 x ^{2} + 5 x + 4}{( 1 + x ) ( 1 + 4 x ^{2} )} d x

Take the partial fraction of

\frac{6 x ^{2} + 5 x + 4}{( 1 + x ) ( 1 + 4 x ^{2} )} : \frac{1}{x + 1} + \frac{2 x + 3}{4 x ^{2} + 1}

= \int \frac{1}{x + 1} + \frac{2 x + 3}{4 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{1}{x + 1} d x + \int \frac{2 x + 3}{4 x ^{2} + 1} d x

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

\int \frac{2 x + 3}{4 x ^{2} + 1} d x = \frac{1}{4} ln 4 x^{2} + 1 + \frac{3}{2} arctan (2 x)

= ln ∣ x + 1 ∣ + \frac{1}{4} ln 4 x^{2} + 1 + \frac{3}{2} arctan (2 x)

Add a constant to the solution

= ln ∣ x + 1 ∣ + \frac{1}{4} ln 4 x^{2} + 1 + \frac{3}{2} arctan (2 x) + C

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What is the integral of (6x^2+5x+4)/((1+x)(1+4x^2)) ?
The integral of (6x^2+5x+4)/((1+x)(1+4x^2)) is ln|x+1|+1/4 ln|4x^2+1|+3/2 arctan(2x)+C