What is the integral of (2x+2x-3)^{35}(2x+1) ?

The integral of (2x+2x-3)^{35}(2x+1) is 1/10656 (36(4x-3)^{37}+185(4x-3)^{36})+C

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integral of (2x+2x-3)^{35}(2x+1)

\int (2 x + 2 x - 3)^{35} (2 x + 1) d x

\frac{1}{10656} (36 (4 x - 3)^{37} + 185 (4 x - 3)^{36}) + C

Solution steps

\int (2 x + 2 x - 3)^{35} (2 x + 1) d x

Apply u-substitution

: \int \frac{u ^{35} ( u + 5 )}{8} d u

= \int \frac{u ^{35} ( u + 5 )}{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^{35} (u + 5) d u

Expand

u^{35} (u + 5) : u^{36} + 5 u^{35}

= \frac{1}{8} \cdot \int u^{36} + 5 u^{35} 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}{8} (\int u^{36} d u + \int 5 u^{35} d u)

\int u^{36} d u = \frac{u ^{37}}{37}

\int 5 u^{35} d u = \frac{5 u ^{36}}{36}

= \frac{1}{8} (\frac{u ^{37}}{37} + \frac{5 u ^{36}}{36})

Substitute back

u = 2 x + 2 x - 3

= \frac{1}{8} (\frac{( 2 x + 2 x - 3 ) ^{37}}{37} + \frac{5 ( 2 x + 2 x - 3 ) ^{36}}{36})

Simplify

\frac{1}{8} (\frac{( 2 x + 2 x - 3 ) ^{37}}{37} + \frac{5 ( 2 x + 2 x - 3 ) ^{36}}{36}) : \frac{1}{10656} (36 (4 x - 3)^{37} + 185 (4 x - 3)^{36})

= \frac{1}{10656} (36 (4 x - 3)^{37} + 185 (4 x - 3)^{36})

Add a constant to the solution

= \frac{1}{10656} (36 (4 x - 3)^{37} + 185 (4 x - 3)^{36}) + C

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What is the integral of (2x+2x-3)^{35}(2x+1) ?
The integral of (2x+2x-3)^{35}(2x+1) is 1/10656 (36(4x-3)^{37}+185(4x-3)^{36})+C