What is the integral of (16)/(x(x^2+4)^2) ?

The integral of (16)/(x(x^2+4)^2) is ln(x)-1/2 ln|x^2+4|+2/(x^2+4)+C

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integral of (16)/(x(x^2+4)^2)

\int \frac{16}{x ( x ^{2} + 4 ) ^{2}} d x

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

Solution steps

\int \frac{16}{x ( x ^{2} + 4 ) ^{2}} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{x ( x ^{2} + 4 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{2 ( u - 4 ) u ^{2}} d u

= 16 \cdot \int \frac{1}{2 ( u - 4 ) u ^{2}} d u

Take the constant out:

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

= 16 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 4 ) u ^{2}} d u

Take the partial fraction of

\frac{1}{( u - 4 ) u ^{2}} : \frac{1}{16 ( u - 4 )} - \frac{1}{16 u} - \frac{1}{4 u ^{2}}

= 16 \cdot \frac{1}{2} \cdot \int \frac{1}{16 ( u - 4 )} - \frac{1}{16 u} - \frac{1}{4 u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 \cdot \frac{1}{2} (\int \frac{1}{16 ( u - 4 )} d u - \int \frac{1}{16 u} d u - \int \frac{1}{4 u ^{2}} d u)

\int \frac{1}{16 ( u - 4 )} d u = \frac{1}{16} ln ∣ u - 4 ∣

\int \frac{1}{16 u} d u = \frac{1}{16} ln ∣ u ∣

\int \frac{1}{4 u ^{2}} d u = - \frac{1}{4 u}

= 16 \cdot \frac{1}{2} (\frac{1}{16} ln ∣ u - 4 ∣ - \frac{1}{16} ln ∣ u ∣ - (- \frac{1}{4 u}))

Substitute back

u = x^{2} + 4

= 16 \cdot \frac{1}{2} (\frac{1}{16} ln x^{2} + 4 - 4 - \frac{1}{16} ln x^{2} + 4 - (- \frac{1}{4 ( x ^{2} + 4 )}))

Simplify

16 \cdot \frac{1}{2} (\frac{1}{16} ln x^{2} + 4 - 4 - \frac{1}{16} ln x^{2} + 4 - (- \frac{1}{4 ( x ^{2} + 4 )})) : ln (x) - \frac{1}{2} ln x^{2} + 4 + \frac{2}{x ^{2} + 4}

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

Add a constant to the solution

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

\frac{d}{d x} ((x sin (2 x))^{\frac{3}{2}})

\int \frac{( x + 16 )}{( x ^{2} + 2 x - 8 )} d x

\frac{\partial}{\partial α} (K (α) (α)^{α} L^{1 - α})

y = 11 x^{2}, (- 2, 44)

\int \frac{1}{r ( 2 - r ^{2} )} d r

What is the integral of (16)/(x(x^2+4)^2) ?
The integral of (16)/(x(x^2+4)^2) is ln(x)-1/2 ln|x^2+4|+2/(x^2+4)+C