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

The integral of (x^3)/(1+4x^2) is 1/32 (4x^2+1-ln|4x^2+1|)+C

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

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

\frac{1}{32} (4 x^{2} + 1 - ln 4 x^{2} + 1) + C

Solution steps

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

Long division

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

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

Apply u-substitution

: \int \frac{u - 1}{32 u} d u

= \int \frac{u - 1}{32 u} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int \frac{u - 1}{u} d u

Expand

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

= \frac{1}{32} \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}{32} (\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}{32} (u - ln ∣ u ∣)

Substitute back

u = 4 x^{2} + 1

= \frac{1}{32} (4 x^{2} + 1 - ln 4 x^{2} + 1)

Add a constant to the solution

= \frac{1}{32} (4 x^{2} + 1 - ln 4 x^{2} + 1) + C

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What is the integral of (x^3)/(1+4x^2) ?
The integral of (x^3)/(1+4x^2) is 1/32 (4x^2+1-ln|4x^2+1|)+C