What is the integral of t(t^2+1)^8 ?

The integral of t(t^2+1)^8 is 1/18 (t^2+1)^9+C

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integral of t(t^2+1)^8

\int t (t^{2} + 1)^{8} d t

\frac{1}{18} (t^{2} + 1)^{9} + C

Solution steps

\int t (t^{2} + 1)^{8} d t

Apply u-substitution

= \int \frac{u ^{8}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u^{8} d u

Apply the Power Rule

= \frac{1}{2} \cdot \frac{u ^{9}}{9}

Substitute back

u = t^{2} + 1

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

Simplify

\frac{1}{2} \cdot \frac{( t ^{2} + 1 ) ^{9}}{9} : \frac{1}{18} (t^{2} + 1)^{9}

= \frac{1}{18} (t^{2} + 1)^{9}

Add a constant to the solution

= \frac{1}{18} (t^{2} + 1)^{9} + C

What is the integral of t(t^2+1)^8 ?
The integral of t(t^2+1)^8 is 1/18 (t^2+1)^9+C