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

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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

\int_{- 1}^{0} y + 1 d y

y^{^{'}} = 2 t, y (0) = 0

\int \frac{1}{486} (81 - x^{2}) d x

n = 1 \sum \infty \frac{1}{n ^{2} + 4}

\frac{\partial}{\partial x} (\frac{( 3 x - 2 y )}{x ^{2} + 3 y})

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