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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \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}{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}{2} (u - ln ∣ u ∣)

Substitute back

u = x^{2} + 1

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

Add a constant to the solution

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

\int 2500 - x^{2} d x

\frac{d}{d x} ((2 x^{2} - 1)^{3} \cdot (1 - x^{2})^{- 3})

\frac{d}{d x} (2 + x^{2})

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

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