What is the integral of 3x(4x-1)^5 ?

The integral of 3x(4x-1)^5 is 1/224 (6(4x-1)^7+7(4x-1)^6)+C

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integral of 3x(4x-1)^5

\int 3 x (4 x - 1)^{5} d x

\frac{1}{224} (6 (4 x - 1)^{7} + 7 (4 x - 1)^{6}) + C

Solution steps

\int 3 x (4 x - 1)^{5} d x

Take the constant out:

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

= 3 \cdot \int x (4 x - 1)^{5} d x

Apply u-substitution

= 3 \cdot \int \frac{u ^{5} ( u + 1 )}{16} d u

Take the constant out:

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

= 3 \cdot \frac{1}{16} \cdot \int u^{5} (u + 1) d u

Expand

u^{5} (u + 1) : u^{6} + u^{5}

= 3 \cdot \frac{1}{16} \cdot \int u^{6} + u^{5} 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

= 3 \cdot \frac{1}{16} (\int u^{6} d u + \int u^{5} d u)

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

\int u^{5} d u = \frac{u ^{6}}{6}

= 3 \cdot \frac{1}{16} (\frac{u ^{7}}{7} + \frac{u ^{6}}{6})

Substitute back

u = 4 x - 1

= 3 \cdot \frac{1}{16} (\frac{( 4 x - 1 ) ^{7}}{7} + \frac{( 4 x - 1 ) ^{6}}{6})

Simplify

3 \cdot \frac{1}{16} (\frac{( 4 x - 1 ) ^{7}}{7} + \frac{( 4 x - 1 ) ^{6}}{6}) : \frac{1}{224} (6 (4 x - 1)^{7} + 7 (4 x - 1)^{6})

= \frac{1}{224} (6 (4 x - 1)^{7} + 7 (4 x - 1)^{6})

Add a constant to the solution

= \frac{1}{224} (6 (4 x - 1)^{7} + 7 (4 x - 1)^{6}) + C

What is the integral of 3x(4x-1)^5 ?
The integral of 3x(4x-1)^5 is 1/224 (6(4x-1)^7+7(4x-1)^6)+C