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

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

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

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

x^{2} + 1 - ln x^{2} + 1 + C

Solution steps

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

Apply u-substitution

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

= 2 \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

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

Expand

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

= 2 \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

= 2 \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 ∣

= 2 \cdot \frac{1}{2} (u - ln ∣ u ∣)

Substitute back

u = x^{2} + 1

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

Simplify

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

= x^{2} + 1 - ln x^{2} + 1

Add a constant to the solution

= x^{2} + 1 - ln x^{2} + 1 + C

x \to 0 lim (x \cdot ln (x))

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

\frac{d}{d u} (\frac{u}{u + v})

\frac{\partial}{\partial x} (8 yz ln (x))

x \to 0 lim (x^{2} - 5)

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