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

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

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

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

ln ∣ x - 1 ∣ - \frac{5}{x - 1} + C

Solution steps

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

Complete the square

x^{2} - 2 x + 1 : (x - 1)^{2}

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

Apply u-substitution

: \int \frac{u + 5}{u ^{2}} d u

= \int \frac{u + 5}{u ^{2}} d u

Expand

\frac{u + 5}{u ^{2}} : \frac{1}{u} + \frac{5}{u ^{2}}

= \int \frac{1}{u} + \frac{5}{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

= \int \frac{1}{u} d u + \int \frac{5}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{5}{u ^{2}} d u = - \frac{5}{u}

= ln ∣ u ∣ - \frac{5}{u}

Substitute back

u = x - 1

= ln ∣ x - 1 ∣ - \frac{5}{x - 1}

Add a constant to the solution

= ln ∣ x - 1 ∣ - \frac{5}{x - 1} + C

\frac{d}{d x} (arctan (\frac{x ^{2}}{y}))

11 x^{\frac{1}{2}} + x^{\frac{3}{2}}

\int_{1}^{e} ln (5 x) d x

x + 1

\frac{d}{d x} (∣ x ∣^{a})

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