What is the integral of (x^3)/(9x^2+1) ?

The integral of (x^3)/(9x^2+1) is 1/162 (9x^2+1-ln|9x^2+1|)+C

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integral of (x^3)/(9x^2+1)

\int \frac{x ^{3}}{9 x ^{2} + 1} d x

\frac{1}{162} (9 x^{2} + 1 - ln 9 x^{2} + 1) + C

Solution steps

\int \frac{x ^{3}}{9 x ^{2} + 1} d x

Long division

\frac{x ^{3}}{9 x ^{2} + 1} : \frac{x}{9} - \frac{x}{9 ( 9 x ^{2} + 1 )}

= \int \frac{x}{9} - \frac{x}{9 ( 9 x ^{2} + 1 )} d x

Apply u-substitution

= \int \frac{u - 1}{162 u} d u

Take the constant out:

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

= \frac{1}{162} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \frac{1}{162} \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{162} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{162} (u - ln ∣ u ∣)

Substitute back

u = 9 x^{2} + 1

= \frac{1}{162} (9 x^{2} + 1 - ln 9 x^{2} + 1)

Add a constant to the solution

= \frac{1}{162} (9 x^{2} + 1 - ln 9 x^{2} + 1) + C

What is the integral of (x^3)/(9x^2+1) ?
The integral of (x^3)/(9x^2+1) is 1/162 (9x^2+1-ln|9x^2+1|)+C