What is the integral from 1 to 2 of (ln(1+x))/(x^2) ?

The integral from 1 to 2 of (ln(1+x))/(x^2) is 3ln(2)-3/2 ln(3)

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integral from 1 to 2 of (ln(1+x))/(x^2)

\int_{1}^{2} \frac{ln ( 1 + x )}{x ^{2}} d x

3 ln (2) - \frac{3}{2} ln (3)

Decimal

0.43152 \dots

Solution steps

\int_{1}^{2} \frac{ln ( 1 + x )}{x ^{2}} d x

Apply Integration By Parts

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

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

\int - \frac{1}{x ( 1 + x )} d x = - ln ∣ x ∣ + ln ∣ x + 1 ∣

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

Simplify

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

Compute the boundaries:

3 ln (2) - \frac{3}{2} ln (3)

= 3 ln (2) - \frac{3}{2} ln (3)

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What is the integral from 1 to 2 of (ln(1+x))/(x^2) ?
The integral from 1 to 2 of (ln(1+x))/(x^2) is 3ln(2)-3/2 ln(3)