What is the integral of x/((4x^2+3)^6) ?

The integral of x/((4x^2+3)^6) is -1/(40(4x^2+3)^5)+C

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integral of x/((4x^2+3)^6)

\int \frac{x}{( 4 x ^{2} + 3 ) ^{6}} d x

- \frac{1}{40 ( 4 x ^{2} + 3 ) ^{5}} + C

Solution steps

\int \frac{x}{( 4 x ^{2} + 3 ) ^{6}} d x

Apply u-substitution

= \int \frac{1}{8 u ^{6}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u ^{6}} d u

Apply the Power Rule

= \frac{1}{8} (- \frac{1}{5 u ^{5}})

Substitute back

u = 4 x^{2} + 3

= \frac{1}{8} (- \frac{1}{5 ( 4 x ^{2} + 3 ) ^{5}})

Simplify

\frac{1}{8} (- \frac{1}{5 ( 4 x ^{2} + 3 ) ^{5}}) : - \frac{1}{40 ( 4 x ^{2} + 3 ) ^{5}}

= - \frac{1}{40 ( 4 x ^{2} + 3 ) ^{5}}

Add a constant to the solution

= - \frac{1}{40 ( 4 x ^{2} + 3 ) ^{5}} + C

t an g e n t \frac{3}{2 3 x + 7}, at x = 6

\frac{d}{d x} (x^{2} arctan (x))

\frac{2 d y}{d x} + 5 y = 4

x \to 0 lim (\frac{1}{3})

4 x y^{^{'}} - (1 + x) y = 2 x y^{2}

What is the integral of x/((4x^2+3)^6) ?
The integral of x/((4x^2+3)^6) is -1/(40(4x^2+3)^5)+C