What is the integral of 1/((2x+2)[ln(x+1)]^2) ?

The integral of 1/((2x+2)[ln(x+1)]^2) is -1/(2ln(x+1))+C

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integral of 1/((2x+2)[ln(x+1)]^2)

\int \frac{1}{( 2 x + 2 ) [ ln ( x + 1 ) ] ^{2}} d x

- \frac{1}{2 ln ( x + 1 )} + C

Solution steps

\int \frac{1}{( 2 x + 2 ) [ ln ( x + 1 ) ] ^{2}} d x

\frac{1}{( 2 x + 2 ) [ ln ( x + 1 ) ] ^{2}} = \frac{1}{2} \frac{1}{( x + 1 ) [ ln ( x + 1 ) ] ^{2}}

= \int \frac{1}{2} \cdot \frac{1}{( x + 1 ) [ ln ( x + 1 ) ] ^{2}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{( x + 1 ) [ ln ( x + 1 ) ] ^{2}} d x

Simplify

= \frac{1}{2} \cdot \int \frac{1}{ln ^{2} ( x + 1 ) ( x + 1 )} d x

Apply u-substitution

: \int \frac{1}{u ln ^{2} ( u )} d u

= \frac{1}{2} \cdot \int \frac{1}{u ln ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{1}{2} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{2} (- \frac{1}{v})

Substitute back

= \frac{1}{2} (- \frac{1}{ln ( x + 1 )})

Simplify

\frac{1}{2} (- \frac{1}{ln ( x + 1 )}) : - \frac{1}{2 ln ( x + 1 )}

= - \frac{1}{2 ln ( x + 1 )}

Add a constant to the solution

= - \frac{1}{2 ln ( x + 1 )} + C

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What is the integral of 1/((2x+2)[ln(x+1)]^2) ?
The integral of 1/((2x+2)[ln(x+1)]^2) is -1/(2ln(x+1))+C