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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int u^{2} + 1 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

= \int u^{2} d u + \int 1 d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= \frac{u ^{3}}{3} + u

Substitute back

u = x^{2} - 1

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

Simplify

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

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

Add a constant to the solution

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

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