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

The integral of x/((a+bx^2)^3) is -1/(4b(a+bx^2)^2)+C

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

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

- \frac{1}{4 b ( a + b x ^{2} ) ^{2}} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{1}{2 b} (- \frac{1}{2 u ^{2}})

Substitute back

u = a + b x^{2}

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

Simplify

\frac{1}{2 b} (- \frac{1}{2 ( a + b x ^{2} ) ^{2}}) : - \frac{1}{4 b ( a + b x ^{2} ) ^{2}}

= - \frac{1}{4 b ( a + b x ^{2} ) ^{2}}

Add a constant to the solution

= - \frac{1}{4 b ( a + b x ^{2} ) ^{2}} + C

y = \frac{x ^{2}}{2} - 3, y = - \frac{x ^{2}}{2} + 2 x

\int_{1}^{9} \frac{1}{x - 7} d x

\frac{d}{d x} (- x e^{(\frac{x + 1}{x - 1})})

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

\frac{d}{d x} (x + 5 e^{x})

What is the integral of x/((a+bx^2)^3) ?
The integral of x/((a+bx^2)^3) is -1/(4b(a+bx^2)^2)+C