What is the integral of (3x)/((2x^2+1)^3) ?

The integral of (3x)/((2x^2+1)^3) is -3/(8(2x^2+1)^2)+C

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integral of (3x)/((2x^2+1)^3)

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

- \frac{3}{8 ( 2 x ^{2} + 1 ) ^{2}} + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{x}{( 2 x ^{2} + 1 ) ^{3}} d x

Apply u-substitution

= 3 \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

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

Apply the Power Rule

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

Substitute back

u = 2 x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

k = 0 \sum \infty x^{k}

t a y l or x^{2} - 4, 0

\frac{d}{d x} (\frac{3 e ^{x}}{5 e ^{x} + 4})

t an g e n t f (x) = - \frac{3}{x ^{2} - 4}, (1, 1)

\frac{d}{d x} (ln (1 + ln (x)))

What is the integral of (3x)/((2x^2+1)^3) ?
The integral of (3x)/((2x^2+1)^3) is -3/(8(2x^2+1)^2)+C