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

The integral of (x^2)/((a+bx^3)^2) is -1/(3b(a+bx^3))+C

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integral of (x^2)/((a+bx^3)^2)

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

- \frac{1}{3 b ( a + b x ^{3} )} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 b u ^{2}} d u

= \int \frac{1}{3 b u ^{2}} d u

Take the constant out:

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

= \frac{1}{3 b} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{3 b} (- \frac{1}{u})

Substitute back

u = a + b x^{3}

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{3 b ( a + b x ^{3} )} + C

x \to 0 + lim ((tan (7 x))^{x})

(2 y^{2} + 3 x) d x + 4 x y d y = 0

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

\int_{- 2}^{2} y^{2} - 4 d y

f (x) = 4 cos (2 x)

What is the integral of (x^2)/((a+bx^3)^2) ?
The integral of (x^2)/((a+bx^3)^2) is -1/(3b(a+bx^3))+C