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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (xln(1+x^2))/(sqrt(1+x^2))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{ln ( 1 + u )}{2 1 + u} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{ln ( 1 + u )}{1 + u} d u

Apply Integration By Parts

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

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

\int \frac{2}{1 + u} d u = 4 1 + u

= \frac{1}{2} (2 ln (1 + u) 1 + u - 4 1 + u)

Substitute back

u = x^{2}

= \frac{1}{2} (2 ln (1 + x^{2}) 1 + x^{2} - 4 1 + x^{2})

Simplify

\frac{1}{2} (2 ln (1 + x^{2}) 1 + x^{2} - 4 1 + x^{2}) : (ln (x^{2} + 1) - 2) x^{2} + 1

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

Add a constant to the solution

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

\int (\frac{2}{3})^{x} d x

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

\int \frac{4 x ^{2}}{x ^{2} + 1} d x

\int x (8 x)^{\frac{3}{2}} + 7 d x

\int (\frac{x ^{2}}{2}) d x

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