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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{4 u ^{3}} d u

= 4 \cdot \int \frac{1}{4 u ^{3}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 4 \cdot \frac{1}{4} (- \frac{1}{2 u ^{2}})

Substitute back

u = 2 x^{2} + 3

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

Simplify

4 \cdot \frac{1}{4} (- \frac{1}{2 ( 2 x ^{2} + 3 ) ^{2}}) : - \frac{1}{2 ( 2 x ^{2} + 3 ) ^{2}}

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

Add a constant to the solution

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

\int \frac{x ^{2} - 3}{x + 1} d x

\int \frac{1}{4 x ^{2} + 1} d x

x \to 1 lim (3 f (x) - 2 h (x))

y = 3^{c o s (2 x)} + e^{3}

\int 7 sin (\frac{θ}{3}) d θ

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