What is the integral of (x-1)/((x^2+4x+5)(x+1)) ?

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

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integral of (x-1)/((x^2+4x+5)(x+1))

\int \frac{x - 1}{( x ^{2} + 4 x + 5 ) ( x + 1 )} d x

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

Solution steps

\int \frac{x - 1}{( x ^{2} + 4 x + 5 ) ( x + 1 )} d x

Complete the square

x^{2} + 4 x + 5 : (x + 2)^{2} + 1

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

Apply u-substitution

: \int \frac{u - 3}{( u ^{2} + 1 ) ( u - 1 )} d u

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

Take the partial fraction of

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

= \int \frac{u + 2}{u ^{2} + 1} - \frac{1}{u - 1} 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

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

\int \frac{u + 2}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1 + 2 arctan (u)

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

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

Substitute back

u = x + 2

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 5 + 2 arctan (x + 2) - ln ∣ x + 1 ∣ + C

\frac{d y}{d x} + x^{2} y = 10 x^{2}

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 cos (4 t)

\int_{0}^{0.6} 250 x d x

\int - 2 sin (2 t) d t

\frac{d}{d x} y = (- 2. 6^{\frac{1}{2}})

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