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

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

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

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

x^{2} + x + C

Solution steps

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

Simplify

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

= \int 2 x + 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 2 x d x + \int 1 d x

\int 2 x d x = x^{2}

\int 1 d x = x

= x^{2} + x

Add a constant to the solution

= x^{2} + x + C

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