What is the integral of (x^2+2x)/(sqrt(x^3+3x^2+1)) ?

The integral of (x^2+2x)/(sqrt(x^3+3x^2+1)) is 2/3 sqrt(x^3+3x^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 (x^2+2x)/(sqrt(x^3+3x^2+1))

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

\frac{2}{3} x^{3} + 3 x^{2} + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{3} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{3} + 3 x^{2} + 1

= \frac{1}{3} \cdot 2 (x^{3} + 3 x^{2} + 1)^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot 2 (x^{3} + 3 x^{2} + 1)^{\frac{1}{2}} : \frac{2}{3} x^{3} + 3 x^{2} + 1

= \frac{2}{3} x^{3} + 3 x^{2} + 1

Add a constant to the solution

= \frac{2}{3} x^{3} + 3 x^{2} + 1 + C

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

f (x) = 2 x^{\frac{1}{2}}

\int 3 sin^{2} (x) cos^{4} (x) d x

\int_{0}^{3} (x - 1) (x - 2) (x - 3) d x

What is the integral of (x^2+2x)/(sqrt(x^3+3x^2+1)) ?
The integral of (x^2+2x)/(sqrt(x^3+3x^2+1)) is 2/3 sqrt(x^3+3x^2+1)+C