What is the integral of (8x^2-4x+7)/(4x^3+x^2+4x+1) ?

The integral of (8x^2-4x+7)/(4x^3+x^2+4x+1) is 2ln|4x+1|-arctan(x)+C

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integral of (8x^2-4x+7)/(4x^3+x^2+4x+1)

\int \frac{8 x ^{2} - 4 x + 7}{4 x ^{3} + x ^{2} + 4 x + 1} d x

2 ln ∣ 4 x + 1 ∣ - arctan (x) + C

Solution steps

\int \frac{8 x ^{2} - 4 x + 7}{4 x ^{3} + x ^{2} + 4 x + 1} d x

Take the partial fraction of

\frac{8 x ^{2} - 4 x + 7}{4 x ^{3} + x ^{2} + 4 x + 1} : \frac{8}{4 x + 1} - \frac{1}{x ^{2} + 1}

= \int \frac{8}{4 x + 1} - \frac{1}{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{8}{4 x + 1} d x - \int \frac{1}{x ^{2} + 1} d x

\int \frac{8}{4 x + 1} d x = 2 ln ∣ 4 x + 1 ∣

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

= 2 ln ∣ 4 x + 1 ∣ - arctan (x)

Add a constant to the solution

= 2 ln ∣ 4 x + 1 ∣ - arctan (x) + C

f (x) = - x^{2} - 4 x + 2, at x = - 1

\frac{d}{d x} (\frac{x ^{2} - 36}{x + 6})

\frac{d}{d x} (3^{x + 10})

f (x) = 2 sin (2 x)

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

What is the integral of (8x^2-4x+7)/(4x^3+x^2+4x+1) ?
The integral of (8x^2-4x+7)/(4x^3+x^2+4x+1) is 2ln|4x+1|-arctan(x)+C