What is the integral of (4x^2-1)^3 ?

The integral of (4x^2-1)^3 is 1/35 x(320x^6-336x^4+140x^2-35)+C

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integral of (4x^2-1)^3

\int (4 x^{2} - 1)^{3} d x

\frac{1}{35} x (320 x^{6} - 336 x^{4} + 140 x^{2} - 35) + C

Solution steps

\int (4 x^{2} - 1)^{3} d x

Apply Integral Substitution

= \int \frac{( u ^{2} - 1 ) ^{3}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int (u^{2} - 1)^{3} d u

Expand

(u^{2} - 1)^{3} : u^{6} - 3 u^{4} + 3 u^{2} - 1

= \frac{1}{2} \cdot \int u^{6} - 3 u^{4} + 3 u^{2} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int u^{6} d u - \int 3 u^{4} d u + \int 3 u^{2} d u - \int 1 d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int 3 u^{4} d u = \frac{3 u ^{5}}{5}

\int 3 u^{2} d u = u^{3}

\int 1 d u = u

= \frac{1}{2} (\frac{u ^{7}}{7} - \frac{3 u ^{5}}{5} + u^{3} - u)

Substitute back

u = 2 x

= \frac{1}{2} (\frac{( 2 x ) ^{7}}{7} - \frac{3 ( 2 x ) ^{5}}{5} + (2 x)^{3} - 2 x)

Simplify

\frac{1}{2} (\frac{( 2 x ) ^{7}}{7} - \frac{3 ( 2 x ) ^{5}}{5} + (2 x)^{3} - 2 x) : \frac{1}{35} x (320 x^{6} - 336 x^{4} + 140 x^{2} - 35)

= \frac{1}{35} x (320 x^{6} - 336 x^{4} + 140 x^{2} - 35)

Add a constant to the solution

= \frac{1}{35} x (320 x^{6} - 336 x^{4} + 140 x^{2} - 35) + C

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What is the integral of (4x^2-1)^3 ?
The integral of (4x^2-1)^3 is 1/35 x(320x^6-336x^4+140x^2-35)+C