What is the integral of (x^2+4)^53x ?

The integral of (x^2+4)^53x is 1/4 (x^2+4)^6+C

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integral of (x^2+4)^53x

\int (x^{2} + 4)^{5} 3 x d x

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

Solution steps

\int (x^{2} + 4)^{5} \cdot 3 x d x

Take the constant out:

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

= 3 \cdot \int (x^{2} + 4)^{5} x 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} + 4

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (5 x^{2} - 3 x y + x yz)

- 2 t^{3}

\int \frac{x}{x ^{2} - 2 x} d x

x \to 0 lim (x^{2} sin (x))

\int - 5 sin (5 t) d t

What is the integral of (x^2+4)^53x ?
The integral of (x^2+4)^53x is 1/4 (x^2+4)^6+C