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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{2 ^{x}}{ln ( 2 )} x d x

\frac{d y}{d x} = \frac{( y ^{2} + x y + x ^{2} )}{x ^{2}}

\int_{0}^{2} 2 d x

f (x) = (x + 2) (e^{- x})

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