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

The integral of (x^3)/(3(x^2+1)) is 1/6 (x^2+1-ln|x^2+1|)+C

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

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

\frac{1}{6} (x^{2} + 1 - ln x^{2} + 1) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u - 1}{2 u} d u

= \frac{1}{3} \cdot \int \frac{u - 1}{2 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int \frac{u - 1}{u} d u

Expand

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

= \frac{1}{3} \cdot \frac{1}{2} \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}{3} \cdot \frac{1}{2} (\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}{3} \cdot \frac{1}{2} (u - ln ∣ u ∣)

Substitute back

u = x^{2} + 1

= \frac{1}{3} \cdot \frac{1}{2} (x^{2} + 1 - ln x^{2} + 1)

Simplify

\frac{1}{3} \cdot \frac{1}{2} (x^{2} + 1 - ln x^{2} + 1) : \frac{1}{6} (x^{2} + 1 - ln x^{2} + 1)

= \frac{1}{6} (x^{2} + 1 - ln x^{2} + 1)

Add a constant to the solution

= \frac{1}{6} (x^{2} + 1 - ln x^{2} + 1) + C

e^{t a n (x)}

f (x) = \frac{x}{x ^{2} + 64}

2 (1 - e^{- t})

n = 1 \sum \infty n (\frac{3}{4})^{n}

f (x) = \frac{sin ( x )}{x ^{2}}

What is the integral of (x^3)/(3(x^2+1)) ?
The integral of (x^3)/(3(x^2+1)) is 1/6 (x^2+1-ln|x^2+1|)+C