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

The integral of 3x(x^2+2)^5 is 1/4 (x^2+2)^6+C

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

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

\frac{1}{4} (x^{2} + 2)^{6} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

= 3 \cdot \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

= 3 \cdot \frac{1}{2} \cdot \int u^{5} d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} + 2

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

Simplify

3 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 2 ) ^{6}}{6} : \frac{1}{4} (x^{2} + 2)^{6}

= \frac{1}{4} (x^{2} + 2)^{6}

Add a constant to the solution

= \frac{1}{4} (x^{2} + 2)^{6} + C

\frac{d}{d x} (\frac{x ^{2} - 3 x}{x + 1})

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

t \frac{d y}{d t} = t^{3} + 15 t^{3} y

- 6 y^{^{'}} + 18 x^{2} = 0

What is the integral of 3x(x^2+2)^5 ?
The integral of 3x(x^2+2)^5 is 1/4 (x^2+2)^6+C