What is the integral of (10x^2)/(8+9x) ?

The integral of (10x^2)/(8+9x) is 10((x^2)/(18)-8/81 x+64/729 ln|9x+8|)+C

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integral of (10x^2)/(8+9x)

\int \frac{10 x ^{2}}{8 + 9 x} d x

10 (\frac{x ^{2}}{18} - \frac{8}{81} x + \frac{64}{729} ln ∣ 9 x + 8 ∣) + C

Solution steps

\int \frac{10 x ^{2}}{8 + 9 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int \frac{x ^{2}}{8 + 9 x} d x

Long division

\frac{x ^{2}}{8 + 9 x} : \frac{x}{9} - \frac{8}{81} + \frac{64}{81 ( 9 x + 8 )}

= 10 \cdot \int \frac{x}{9} - \frac{8}{81} + \frac{64}{81 ( 9 x + 8 )} 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

= 10 (\int \frac{x}{9} d x - \int \frac{8}{81} d x + \int \frac{64}{81 ( 9 x + 8 )} d x)

\int \frac{x}{9} d x = \frac{x ^{2}}{18}

\int \frac{8}{81} d x = \frac{8}{81} x

\int \frac{64}{81 ( 9 x + 8 )} d x = \frac{64}{729} ln ∣ 9 x + 8 ∣

= 10 (\frac{x ^{2}}{18} - \frac{8}{81} x + \frac{64}{729} ln ∣ 9 x + 8 ∣)

Add a constant to the solution

= 10 (\frac{x ^{2}}{18} - \frac{8}{81} x + \frac{64}{729} ln ∣ 9 x + 8 ∣) + C

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What is the integral of (10x^2)/(8+9x) ?
The integral of (10x^2)/(8+9x) is 10((x^2)/(18)-8/81 x+64/729 ln|9x+8|)+C