What is the integral from 0 to 1 of x(x^2+3)^n ?

The integral from 0 to 1 of x(x^2+3)^n is (4^{n+1}-3^{n+1})/(2(n+1))

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integral from 0 to 1 of x(x^2+3)^n

\int_{0}^{1} x (x^{2} + 3)^{n} d x

\frac{4 ^{n + 1} - 3 ^{n + 1}}{2 ( n + 1 )}

Solution steps

\int_{0}^{1} x (x^{2} + 3)^{n} d x

Apply u-substitution

: \int_{3}^{4} \frac{u ^{n}}{2} d u

= \int_{3}^{4} \frac{u ^{n}}{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_{3}^{4} u^{n} d u

Apply the Power Rule

: [\frac{u ^{n + 1}}{n + 1}]_{3}^{4}

= \frac{1}{2} [\frac{u ^{n + 1}}{n + 1}]_{3}^{4}

Compute the boundaries:

\frac{4 ^{n + 1}}{n + 1} - \frac{3 ^{n + 1}}{n + 1}

= \frac{1}{2} (\frac{4 ^{n + 1}}{n + 1} - \frac{3 ^{n + 1}}{n + 1})

Simplify

= \frac{4 ^{n + 1} - 3 ^{n + 1}}{2 ( n + 1 )}

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What is the integral from 0 to 1 of x(x^2+3)^n ?
The integral from 0 to 1 of x(x^2+3)^n is (4^{n+1}-3^{n+1})/(2(n+1))