What is the integral of log_{e}(x+sqrt(1+x^2)) ?

The integral of log_{e}(x+sqrt(1+x^2)) is xln(x+sqrt(1+x^2))-sqrt(1+x^2)+C

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integral of log_{e}(x+sqrt(1+x^2))

\int lo g_{e} (x + 1 + x^{2}) d x

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

Solution steps

\int lo g_{e} (x + 1 + x^{2}) d x

Apply log rule:

lo g_{e} (x) = ln (x)

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

Apply Integration By Parts

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

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

\int \frac{x}{1 + x ^{2}} d x = 1 + x^{2}

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

Add a constant to the solution

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

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What is the integral of log_{e}(x+sqrt(1+x^2)) ?
The integral of log_{e}(x+sqrt(1+x^2)) is xln(x+sqrt(1+x^2))-sqrt(1+x^2)+C