What is the integral of (4x+3)^{25} ?

The integral of (4x+3)^{25} is 1/104 (4x+3)^{26}+C

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integral of (4x+3)^{25}

\int (4 x + 3)^{25} d x

\frac{1}{104} (4 x + 3)^{26} + C

Solution steps

\int (4 x + 3)^{25} d x

Apply u-substitution

: \int \frac{u ^{25}}{4} d u

= \int \frac{u ^{25}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{25} d u

Apply the Power Rule

: \frac{u ^{26}}{26}

= \frac{1}{4} \cdot \frac{u ^{26}}{26}

Substitute back

u = 4 x + 3

= \frac{1}{4} \cdot \frac{( 4 x + 3 ) ^{26}}{26}

Simplify

\frac{1}{4} \cdot \frac{( 4 x + 3 ) ^{26}}{26} : \frac{1}{104} (4 x + 3)^{26}

= \frac{1}{104} (4 x + 3)^{26}

Add a constant to the solution

= \frac{1}{104} (4 x + 3)^{26} + C

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What is the integral of (4x+3)^{25} ?
The integral of (4x+3)^{25} is 1/104 (4x+3)^{26}+C