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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Add a constant to the solution

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

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