What is the integral of (t^2+2t+1)^4(t+1) ?

The integral of (t^2+2t+1)^4(t+1) is ((t+1)^{10})/(10)+C

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integral of (t^2+2t+1)^4(t+1)

\int (t^{2} + 2 t + 1)^{4} (t + 1) d t

\frac{( t + 1 ) ^{10}}{10} + C

Solution steps

\int (t^{2} + 2 t + 1)^{4} (t + 1) d t

Apply u-substitution

: \int u^{9} d u

= \int u^{9} d u

Apply the Power Rule

: \frac{u ^{10}}{10}

= \frac{u ^{10}}{10}

Substitute back

u = t + 1

= \frac{( t + 1 ) ^{10}}{10}

Add a constant to the solution

= \frac{( t + 1 ) ^{10}}{10} + C

\int_{0}^{5} ∣ 9 - 3 t ∣ d t

y = x, y = x^{2}, y = - x + 1

k = 1 \sum \infty k^{k} x^{k}

\frac{d}{d x} (\frac{2 x + 1}{2 x})

x^{2} - 2 x, - x^{2} + 4 x

What is the integral of (t^2+2t+1)^4(t+1) ?
The integral of (t^2+2t+1)^4(t+1) is ((t+1)^{10})/(10)+C