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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 3 x^{2} - 4 x, at x = - 1

\frac{d}{d y} (x - 2 y)

\frac{s + 2}{( s + 1 ) ^{2}}

f (x) = 5 x^{9} x^{6}, at x = 2

y^{^{'}} = \frac{y ^{4} - 3}{4 x y ^{3}}

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