Answered: "Evaluate the integral. ∫(x2+4)x(x2+1)2"

tripiverded9

Answered question

2021-12-12

Evaluate the integral.

\int \frac{(x^{2} + 4)}{x {(x^{2} + 1)}^{2}}

Thomas Lynn

Beginner2021-12-13Added 28 answers

Step 1
The integral is given as,

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

.
After simplify the given integral ,

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

Solving the each part of integral ,
Substitute

u = x^{2} + 1

.
du=2x dx.

\int \frac{x}{x^{2} + 1} d x = \frac{1}{2} \int \frac{d u}{u}

= \frac{\ln u}{2}

Step 2
Replace u with

x^{2} + 1

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

Similarly, substitute

u = x^{2} + 1

.
du=2xdx.

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

= \frac{- 3}{2 u}

Replace u with

x^{2} + 1

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

Step 3
Therefore the solution of the integral is,

\int \frac{(x^{2} + 4)}{x {(x^{2} + 1)}^{2}} d x = - 2 \ln (x^{2} + 1) - \frac{3}{2 (x^{2} + 1)} + 4 \ln x + C

Heather Fulton

Beginner2021-12-14Added 31 answers

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

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

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

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

= \frac{1}{2} \int \frac{1}{u} d u

\int \frac{1}{u} d u

= \ln (u)

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

= \frac{\ln (u)}{2}

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

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

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

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

= - \frac{1}{u}

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

= - \frac{1}{2 u}

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

\int \frac{1}{x} d x

= \ln (x)

- 4 \int \frac{x}{x^{2} + 1} d x - 3 \int \frac{x}{{(x^{2} + 1)}^{2}} d x + 4 \int \frac{1}{x} d x

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