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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

: u ln (u) - \int 1 d u

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

\int 1 d u = u

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

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int x e^{1 + x} d x

\int \frac{5 x ^{2} - x + 1}{16 x ^{4} - 1} d x

\int p (p + 6)^{4} d p

\int sin (v) d v

\int \frac{x ^{3} + 4 x + 3}{x ^{2} + 4} d x

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