What is the integral of 7/((2x+1)[1+ln(2x+1)]^2) ?

The integral of 7/((2x+1)[1+ln(2x+1)]^2) is -7/(2(1+ln(2x+1)))+C

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integral of 7/((2x+1)[1+ln(2x+1)]^2)

\int \frac{7}{( 2 x + 1 ) [ 1 + ln ( 2 x + 1 ) ] ^{2}} d x

- \frac{7}{2 ( 1 + ln ( 2 x + 1 ) )} + C

Solution steps

\int \frac{7}{( 2 x + 1 ) [ 1 + ln ( 2 x + 1 ) ] ^{2}} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{( 2 x + 1 ) [ 1 + ln ( 2 x + 1 ) ] ^{2}} d x

Simplify

(1 + ln (2 x + 1))^{2} : (1 + ln (2 x + 1)) (1 + ln (2 x + 1))

= 7 \cdot \int \frac{1}{( 2 x + 1 ) ( 1 + ln ( 2 x + 1 ) ) ( 1 + ln ( 2 x + 1 ) )} d x

(2 x + 1) (1 + ln (2 x + 1)) (1 + ln (2 x + 1)) = (2 x + 1) (1 + ln (2 x + 1))^{2}

= 7 \cdot \int \frac{1}{( 2 x + 1 ) ( 1 + ln ( 2 x + 1 ) ) ^{2}} d x

Apply u-substitution

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

= 7 \cdot \int \frac{1}{2 u ( 1 + ln ( u ) ) ^{2}} d u

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int \frac{1}{u ( 1 + ln ( u ) ) ^{2}} d u

Apply u-substitution

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

= 7 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 7 \cdot \frac{1}{2} (- \frac{1}{v})

Substitute back

= 7 \cdot \frac{1}{2} (- \frac{1}{1 + ln ( 2 x + 1 )})

Simplify

7 \cdot \frac{1}{2} (- \frac{1}{1 + ln ( 2 x + 1 )}) : - \frac{7}{2 ( 1 + ln ( 2 x + 1 ) )}

= - \frac{7}{2 ( 1 + ln ( 2 x + 1 ) )}

Add a constant to the solution

= - \frac{7}{2 ( 1 + ln ( 2 x + 1 ) )} + C

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What is the integral of 7/((2x+1)[1+ln(2x+1)]^2) ?
The integral of 7/((2x+1)[1+ln(2x+1)]^2) is -7/(2(1+ln(2x+1)))+C