What is the integral of 16x(3-4x)^{40} ?

The integral of 16x(3-4x)^{40} is 1/42 (3-4x)^{42}-3/41 (3-4x)^{41}+C

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integral of 16x(3-4x)^{40}

\int 16 x (3 - 4 x)^{40} d x

\frac{1}{42} (3 - 4 x)^{42} - \frac{3}{41} (3 - 4 x)^{41} + C

Solution steps

\int 16 x (3 - 4 x)^{40} d x

Take the constant out:

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

= 16 \cdot \int x (3 - 4 x)^{40} d x

Apply u-substitution

: \int \frac{u ^{40} ( u - 3 )}{16} d u

= 16 \cdot \int \frac{u ^{40} ( u - 3 )}{16} d u

Take the constant out:

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

= 16 \cdot \frac{1}{16} \cdot \int u^{40} (u - 3) d u

Expand

u^{40} (u - 3) : u^{41} - 3 u^{40}

= 16 \cdot \frac{1}{16} \cdot \int u^{41} - 3 u^{40} 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

= 16 \cdot \frac{1}{16} (\int u^{41} d u - \int 3 u^{40} d u)

\int u^{41} d u = \frac{u ^{42}}{42}

\int 3 u^{40} d u = \frac{3 u ^{41}}{41}

= 16 \cdot \frac{1}{16} (\frac{u ^{42}}{42} - \frac{3 u ^{41}}{41})

Substitute back

u = 3 - 4 x

= 16 \cdot \frac{1}{16} (\frac{( 3 - 4 x ) ^{42}}{42} - \frac{3 ( 3 - 4 x ) ^{41}}{41})

Simplify

16 \cdot \frac{1}{16} (\frac{( 3 - 4 x ) ^{42}}{42} - \frac{3 ( 3 - 4 x ) ^{41}}{41}) : \frac{1}{42} (3 - 4 x)^{42} - \frac{3}{41} (3 - 4 x)^{41}

= \frac{1}{42} (3 - 4 x)^{42} - \frac{3}{41} (3 - 4 x)^{41}

Add a constant to the solution

= \frac{1}{42} (3 - 4 x)^{42} - \frac{3}{41} (3 - 4 x)^{41} + C

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What is the integral of 16x(3-4x)^{40} ?
The integral of 16x(3-4x)^{40} is 1/42 (3-4x)^{42}-3/41 (3-4x)^{41}+C