What is the integral of 8x(4x^2+3)^2 ?

The integral of 8x(4x^2+3)^2 is 1/3 (4x^2+3)^3+C

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integral of 8x(4x^2+3)^2

\int 8 x (4 x^{2} + 3)^{2} d x

\frac{1}{3} (4 x^{2} + 3)^{3} + C

Solution steps

\int 8 x (4 x^{2} + 3)^{2} d x

Take the constant out:

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

= 8 \cdot \int x (4 x^{2} + 3)^{2} d x

Apply u-substitution

: \int \frac{u ^{2}}{8} d u

= 8 \cdot \int \frac{u ^{2}}{8} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= 8 \cdot \frac{1}{8} \cdot \frac{u ^{3}}{3}

Substitute back

u = 4 x^{2} + 3

= 8 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} + 3 ) ^{3}}{3}

Simplify

8 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} + 3 ) ^{3}}{3} : \frac{1}{3} (4 x^{2} + 3)^{3}

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

Add a constant to the solution

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

f (x) = \frac{5}{x}, at x = 4

(lo g_{10} (x))^{^{'}}

63 e^{6 t}

\frac{\partial}{\partial x} (x^{4} - x^{2} y^{3})

6 - 5 x^{3}

What is the integral of 8x(4x^2+3)^2 ?
The integral of 8x(4x^2+3)^2 is 1/3 (4x^2+3)^3+C