What is the integral of (x^2-4)^{3/2} ?

The integral of (x^2-4)^{3/2} is

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integral of (x^2-4)^{3/2}

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

x (x^{2} - 4)^{\frac{3}{2}} - 48 - \frac{1}{32} x (x^{2} - 4)^{\frac{1}{2}} + 4 ln \frac{( x ^{2} - 4 ) ^{\frac{1}{2}} + x}{2} + \frac{1}{64} x^{3} (x^{2} - 4)^{\frac{1}{2}} + C

Solution steps

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

Apply Integration By Parts

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

\int 3 x^{2} (x^{2} - 4)^{\frac{1}{2}} d x = 48 - \frac{1}{32} x (x^{2} - 4)^{\frac{1}{2}} + 4 ln \frac{( x ^{2} - 4 ) ^{\frac{1}{2}} + x}{2} + \frac{1}{64} x^{3} (x^{2} - 4)^{\frac{1}{2}}

= x (x^{2} - 4)^{\frac{3}{2}} - 48 - \frac{1}{32} x (x^{2} - 4)^{\frac{1}{2}} + 4 ln \frac{( x ^{2} - 4 ) ^{\frac{1}{2}} + x}{2} + \frac{1}{64} x^{3} (x^{2} - 4)^{\frac{1}{2}}

Add a constant to the solution

= x (x^{2} - 4)^{\frac{3}{2}} - 48 - \frac{1}{32} x (x^{2} - 4)^{\frac{1}{2}} + 4 ln \frac{( x ^{2} - 4 ) ^{\frac{1}{2}} + x}{2} + \frac{1}{64} x^{3} (x^{2} - 4)^{\frac{1}{2}} + C