What is the integral of (t^3)/(sqrt(t^2-9)) ?

The integral of (t^3)/(sqrt(t^2-9)) is 1/3 (t^2-9)^{3/2}+9sqrt(t^2-9)+C

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integral of (t^3)/(sqrt(t^2-9))

\int \frac{t ^{3}}{t ^{2} - 9} d t

\frac{1}{3} (t^{2} - 9)^{\frac{3}{2}} + 9 t^{2} - 9 + C

Solution steps

\int \frac{t ^{3}}{t ^{2} - 9} d t

Apply u-substitution

= \int u^{2} + 9 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 9 d u

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

\int 9 d u = 9 u

= \frac{u ^{3}}{3} + 9 u

Substitute back

u = t^{2} - 9

= \frac{( t ^{2} - 9 ) ^{3}}{3} + 9 t^{2} - 9

Simplify

\frac{( t ^{2} - 9 ) ^{3}}{3} + 9 t^{2} - 9 : \frac{1}{3} (t^{2} - 9)^{\frac{3}{2}} + 9 t^{2} - 9

= \frac{1}{3} (t^{2} - 9)^{\frac{3}{2}} + 9 t^{2} - 9

Add a constant to the solution

= \frac{1}{3} (t^{2} - 9)^{\frac{3}{2}} + 9 t^{2} - 9 + C

What is the integral of (t^3)/(sqrt(t^2-9)) ?
The integral of (t^3)/(sqrt(t^2-9)) is 1/3 (t^2-9)^{3/2}+9sqrt(t^2-9)+C