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

The integral of (x^3-x)/(sqrt(x^2+1)) is 1/3 (x^2+1)^{3/2}-2sqrt(x^2+1)+C

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (x^3-x)/(sqrt(x^2+1)) ?
The integral of (x^3-x)/(sqrt(x^2+1)) is 1/3 (x^2+1)^{3/2}-2sqrt(x^2+1)+C