What is the integral of (a+bx^2)/(sqrt(3ax+bx^3)) ?

The integral of (a+bx^2)/(sqrt(3ax+bx^3)) is 2/3 sqrt(3ax+bx^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (a+bx^2)/(sqrt(3ax+bx^3))

\int \frac{a + b x ^{2}}{3 a x + b x ^{3}} d x

\frac{2}{3} 3 a x + b x^{3} + C

Solution steps

\int \frac{a + b x ^{2}}{3 a x + b x ^{3}} 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 = 3 a x + b x^{3}

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

Simplify

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

= \frac{2}{3} 3 a x + b x^{3}

Add a constant to the solution

= \frac{2}{3} 3 a x + b x^{3} + C

\frac{d}{d x} (arccos (3 x))

t \frac{d y}{d t} + 21 y = \frac{1}{t}

\int_{0}^{x} (t^{3} - t)^{5} d t

10 + \frac{50 x}{2 x ^{2} + 9}

\int e^{x} sin (5 x) d x

What is the integral of (a+bx^2)/(sqrt(3ax+bx^3)) ?
The integral of (a+bx^2)/(sqrt(3ax+bx^3)) is 2/3 sqrt(3ax+bx^3)+C