What is the integral of ([4-ln(x+3)]^3)/(x+3) ?

The integral of ([4-ln(x+3)]^3)/(x+3) is -((4-ln(x+3))^4)/4+C

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integral of ([4-ln(x+3)]^3)/(x+3)

\int \frac{[ 4 - ln ( x + 3 ) ] ^{3}}{x + 3} d x

- \frac{( 4 - ln ( x + 3 ) ) ^{4}}{4} + C

Solution steps

\int \frac{[ 4 - ln ( x + 3 ) ] ^{3}}{x + 3} d x

Apply u-substitution

: \int - u^{3} d u

= \int - u^{3} d u

Take the constant out:

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

= - \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= - \frac{u ^{4}}{4}

Substitute back

u = 4 - ln (x + 3)

= - \frac{( 4 - ln ( x + 3 ) ) ^{4}}{4}

Add a constant to the solution

= - \frac{( 4 - ln ( x + 3 ) ) ^{4}}{4} + C

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What is the integral of ([4-ln(x+3)]^3)/(x+3) ?
The integral of ([4-ln(x+3)]^3)/(x+3) is -((4-ln(x+3))^4)/4+C