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

The integral of ((x-2))/((x+1)^2+4) is 1/2 ln|x^2+2x+5|-3/2 arctan(1/2 (x+1))+C

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integral of ((x-2))/((x+1)^2+4)

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

\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{3}{2} arctan (\frac{1}{2} (x + 1)) + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 3}{u ^{2} + 4} d u

= \int \frac{u - 3}{u ^{2} + 4} d u

Expand

\frac{u - 3}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} - \frac{3}{u ^{2} + 4}

Apply the Sum Rule:

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

= \int \frac{u}{u ^{2} + 4} d u - \int \frac{3}{u ^{2} + 4} d u

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

\int \frac{3}{u ^{2} + 4} d u = \frac{3}{2} arctan (\frac{u}{2})

= \frac{1}{2} ln u^{2} + 4 - \frac{3}{2} arctan (\frac{u}{2})

Substitute back

u = x + 1

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

Simplify

\frac{1}{2} ln (x + 1)^{2} + 4 - \frac{3}{2} arctan (\frac{x + 1}{2}) : \frac{1}{2} ln x^{2} + 2 x + 5 - \frac{3}{2} arctan (\frac{1}{2} (x + 1))

= \frac{1}{2} ln x^{2} + 2 x + 5 - \frac{3}{2} arctan (\frac{1}{2} (x + 1))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 2 x + 5 - \frac{3}{2} arctan (\frac{1}{2} (x + 1)) + C

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What is the integral of ((x-2))/((x+1)^2+4) ?
The integral of ((x-2))/((x+1)^2+4) is 1/2 ln|x^2+2x+5|-3/2 arctan(1/2 (x+1))+C