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

The integral of x(4x^2+3)^3 is 1/32 (4x^2+3)^4+C

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

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

\frac{1}{32} (4 x^{2} + 3)^{4} + C

Solution steps

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

Apply u-substitution

= \int \frac{u ^{3}}{8} d u

Take the constant out:

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

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

Apply the Power Rule

= \frac{1}{8} \cdot \frac{u ^{4}}{4}

Substitute back

u = 4 x^{2} + 3

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

Simplify

\frac{1}{8} \cdot \frac{( 4 x ^{2} + 3 ) ^{4}}{4} : \frac{1}{32} (4 x^{2} + 3)^{4}

= \frac{1}{32} (4 x^{2} + 3)^{4}

Add a constant to the solution

= \frac{1}{32} (4 x^{2} + 3)^{4} + C

\frac{d}{d x} (x ln (3 x))

y = (x^{3} + 1) (x^{2} - 2)

y^{^{''}} + 15 y^{^{'}} - 56 y = 0

l a pl a ce t r an s f or m e^{(- 2 (t - 3))}

\int \frac{x ^{4} - 2 x ^{3} + 1}{x ^{2}} d x

What is the integral of x(4x^2+3)^3 ?
The integral of x(4x^2+3)^3 is 1/32 (4x^2+3)^4+C