What is the integral of (2x^3)/(36x^2+1) ?

The integral of (2x^3)/(36x^2+1) is 1/1296 (36x^2+1-ln|36x^2+1|)+C

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integral of (2x^3)/(36x^2+1)

\int \frac{2 x ^{3}}{36 x ^{2} + 1} d x

\frac{1}{1296} (36 x^{2} + 1 - ln 36 x^{2} + 1) + C

Solution steps

\int \frac{2 x ^{3}}{36 x ^{2} + 1} d x

Take the constant out:

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

= 2 \cdot \int \frac{x ^{3}}{36 x ^{2} + 1} d x

Long division

\frac{x ^{3}}{36 x ^{2} + 1} : \frac{x}{36} - \frac{x}{36 ( 36 x ^{2} + 1 )}

= 2 \cdot \int \frac{x}{36} - \frac{x}{36 ( 36 x ^{2} + 1 )} d x

Apply u-substitution

: \int \frac{u - 1}{2592 u} d u

= 2 \cdot \int \frac{u - 1}{2592 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2592} \cdot \int \frac{u - 1}{u} d u

Expand

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

= 2 \cdot \frac{1}{2592} \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

= 2 \cdot \frac{1}{2592} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 \cdot \frac{1}{2592} (u - ln ∣ u ∣)

Substitute back

u = 36 x^{2} + 1

= 2 \cdot \frac{1}{2592} (36 x^{2} + 1 - ln 36 x^{2} + 1)

Simplify

2 \cdot \frac{1}{2592} (36 x^{2} + 1 - ln 36 x^{2} + 1) : \frac{1}{1296} (36 x^{2} + 1 - ln 36 x^{2} + 1)

= \frac{1}{1296} (36 x^{2} + 1 - ln 36 x^{2} + 1)

Add a constant to the solution

= \frac{1}{1296} (36 x^{2} + 1 - ln 36 x^{2} + 1) + C

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What is the integral of (2x^3)/(36x^2+1) ?
The integral of (2x^3)/(36x^2+1) is 1/1296 (36x^2+1-ln|36x^2+1|)+C