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

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

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

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

x ln^{2} (x + 1 + x^{2}) - 2 (ln (x + 1 + x^{2}) 1 + x^{2} - x) + C

Solution steps

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

Simplify

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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