What is the integral of (x^2+3)^5x ?

The integral of (x^2+3)^5x is 1/12 (x^2+3)^6+C

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integral of (x^2+3)^5x

\int (x^{2} + 3)^{5} x d x

\frac{1}{12} (x^{2} + 3)^{6} + C

Solution steps

\int (x^{2} + 3)^{5} x d x

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} + 3

= \frac{1}{2} \cdot \frac{( x ^{2} + 3 ) ^{6}}{6}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 3 ) ^{6}}{6} : \frac{1}{12} (x^{2} + 3)^{6}

= \frac{1}{12} (x^{2} + 3)^{6}

Add a constant to the solution

= \frac{1}{12} (x^{2} + 3)^{6} + C

n = 1 \sum \infty ((- 1)^{n + 1}) (\frac{cos ( nπ )}{n ^{2}})

6 e^{5 t} cos (2 t)

f (x) = x^{\frac{5}{2}} e^{x}

2 x + 5

6 y^{^{''}} + y^{^{'}} - y = 0, y (0) = - 1, y^{^{'}} (0) = 3

What is the integral of (x^2+3)^5x ?
The integral of (x^2+3)^5x is 1/12 (x^2+3)^6+C