What is the integral from x^2 to x^4 of (t+t^2)^2 ?

The integral from x^2 to x^4 of (t+t^2)^2 is (x^{12}-x^6)/3+(x^{16}-x^8)/2+(x^{20}-x^{10})/5

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integral from x^2 to x^4 of (t+t^2)^2

\int_{x^{2}}^{x^{4}} (t + t^{2})^{2} d t

\frac{x ^{12} - x ^{6}}{3} + \frac{x ^{16} - x ^{8}}{2} + \frac{x ^{20} - x ^{10}}{5}

Solution steps

\int_{x^{2}}^{x^{4}} (t + t^{2})^{2} d t

Expand

(t + t^{2})^{2} : t^{2} + 2 t^{3} + t^{4}

= \int_{x^{2}}^{x^{4}} t^{2} + 2 t^{3} + t^{4} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{x^{2}}^{x^{4}} t^{2} d t + \int_{x^{2}}^{x^{4}} 2 t^{3} d t + \int_{x^{2}}^{x^{4}} t^{4} d t

\int_{x^{2}}^{x^{4}} t^{2} d t = \frac{x ^{12} - x ^{6}}{3}

\int_{x^{2}}^{x^{4}} 2 t^{3} d t = \frac{x ^{16} - x ^{8}}{2}

\int_{x^{2}}^{x^{4}} t^{4} d t = \frac{x ^{20} - x ^{10}}{5}

= \frac{x ^{12} - x ^{6}}{3} + \frac{x ^{16} - x ^{8}}{2} + \frac{x ^{20} - x ^{10}}{5}

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What is the integral from x^2 to x^4 of (t+t^2)^2 ?
The integral from x^2 to x^4 of (t+t^2)^2 is (x^{12}-x^6)/3+(x^{16}-x^8)/2+(x^{20}-x^{10})/5