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

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

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

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

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

Solution steps

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

Simplify

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int 4 e^{5 x} d x

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cos (sin (15 e^{x}))

\int \frac{x ^{2}}{7 + 3 x ^{3}} d x

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