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

The integral of x^2sqrt(x^2-1) is 1/4 x^3sqrt(x^2-1)-1/8 xsqrt(x^2-1)-1/8 ln|sqrt(x^2-1)+x|+C

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

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

\frac{1}{4} x^{3} x^{2} - 1 - \frac{1}{8} x x^{2} - 1 - \frac{1}{8} ln x^{2} - 1 + x + C

Solution steps

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

Apply Integration By Parts

= \frac{1}{3} x^{3} x^{2} - 1 - \int \frac{x ^{4}}{3 x ^{2} - 1} d x

\int \frac{x ^{4}}{3 x ^{2} - 1} d x = \frac{1}{12} x^{3} x^{2} - 1 + \frac{1}{8} (x x^{2} - 1 + ln x^{2} - 1 + x)

= \frac{1}{3} x^{3} x^{2} - 1 - (\frac{1}{12} x^{3} x^{2} - 1 + \frac{1}{8} (x x^{2} - 1 + ln x^{2} - 1 + x))

Simplify

\frac{1}{3} x^{3} x^{2} - 1 - (\frac{1}{12} x^{3} x^{2} - 1 + \frac{1}{8} (x x^{2} - 1 + ln x^{2} - 1 + x)) : \frac{1}{4} x^{3} x^{2} - 1 - \frac{1}{8} x x^{2} - 1 - \frac{1}{8} ln x^{2} - 1 + x

= \frac{1}{4} x^{3} x^{2} - 1 - \frac{1}{8} x x^{2} - 1 - \frac{1}{8} ln x^{2} - 1 + x

Add a constant to the solution

= \frac{1}{4} x^{3} x^{2} - 1 - \frac{1}{8} x x^{2} - 1 - \frac{1}{8} ln x^{2} - 1 + x + C

What is the integral of x^2sqrt(x^2-1) ?
The integral of x^2sqrt(x^2-1) is 1/4 x^3sqrt(x^2-1)-1/8 xsqrt(x^2-1)-1/8 ln|sqrt(x^2-1)+x|+C