What is the integral of (sqrt(t^9+4t))(9t^8+4) ?

The integral of (sqrt(t^9+4t))(9t^8+4) is 2/3 (t^9+4t)^{3/2}+C

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integral of (sqrt(t^9+4t))(9t^8+4)

\int (t^{9} + 4 t) (9 t^{8} + 4) d t

\frac{2}{3} (t^{9} + 4 t)^{\frac{3}{2}} + C

Solution steps

\int t^{9} + 4 t (9 t^{8} + 4) d t

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = t^{9} + 4 t

= \frac{2}{3} (t^{9} + 4 t)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (t^{9} + 4 t)^{\frac{3}{2}} + C

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What is the integral of (sqrt(t^9+4t))(9t^8+4) ?
The integral of (sqrt(t^9+4t))(9t^8+4) is 2/3 (t^9+4t)^{3/2}+C