What is the integral of 0^{13}x+1(x+1)^2 ?

The integral of 0^{13}x+1(x+1)^2 is (x^3)/3+x^2+x+C

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integral of 0^{13}x+1(x+1)^2

integral

0^{13} x + 1 (x + 1)^{2}

\frac{x ^{3}}{3} + x^{2} + x + C

Solution steps

\int 0^{13} x + 1 \cdot (x + 1)^{2} d x

Simplify

= \int (x + 1)^{2} d x

Expand

(x + 1)^{2} : x^{2} + 2 x + 1

= \int x^{2} + 2 x + 1 d x

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} d x + \int 2 x d x + \int 1 d x

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

\int 2 x d x = x^{2}

\int 1 d x = x

= \frac{x ^{3}}{3} + x^{2} + x

Add a constant to the solution

= \frac{x ^{3}}{3} + x^{2} + x + C

What is the integral of 0^{13}x+1(x+1)^2 ?
The integral of 0^{13}x+1(x+1)^2 is (x^3)/3+x^2+x+C