What is the integral of (x^2+2)^42x ?

The integral of (x^2+2)^42x is 1/5 (x^2+2)^5+C

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integral of (x^2+2)^42x

\int (x^{2} + 2)^{4} 2 x d x

\frac{1}{5} (x^{2} + 2)^{5} + C

Solution steps

\int (x^{2} + 2)^{4} \cdot 2 x d x

Take the constant out:

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

= 2 \cdot \int (x^{2} + 2)^{4} x d x

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

= 2 \cdot \int \frac{u ^{4}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{2} + 2

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 2 ) ^{5}}{5}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 2 ) ^{5}}{5} : \frac{1}{5} (x^{2} + 2)^{5}

= \frac{1}{5} (x^{2} + 2)^{5}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 2)^{5} + C

4 y^{^{''}} + 4 y^{^{'}} + 8 y = 0

x \to 0 lim (cos (9 x))

n = 1 \sum \infty n 2^{- n}

\int x y e^{y^{2} x} d y

\frac{3}{e ^{x} + e ^{- x}}

What is the integral of (x^2+2)^42x ?
The integral of (x^2+2)^42x is 1/5 (x^2+2)^5+C