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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

= \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

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

\int x \cdot 5^{- x^{2}} d x

\int \frac{3}{1 - 3 x - x ^{2}} d x

\int \frac{28 t + 28}{14 t ^{2} + 28 t - 9} d t

\int \frac{e ^{x}}{x + 2} d x

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