What is the integral of (7x+3)/(x^2-2x+1) ?

The integral of (7x+3)/(x^2-2x+1) is 7ln|x-1|-(10)/(x-1)+C

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integral of (7x+3)/(x^2-2x+1)

\int \frac{7 x + 3}{x ^{2} - 2 x + 1} d x

7 ln ∣ x - 1 ∣ - \frac{10}{x - 1} + C

Solution steps

\int \frac{7 x + 3}{x ^{2} - 2 x + 1} d x

Expand

\frac{7 x + 3}{x ^{2} - 2 x + 1} : \frac{7 x}{x ^{2} - 2 x + 1} + \frac{3}{x ^{2} - 2 x + 1}

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{7 x}{x ^{2} - 2 x + 1} d x + \int \frac{3}{x ^{2} - 2 x + 1} d x

\int \frac{7 x}{x ^{2} - 2 x + 1} d x = 7 (ln ∣ x - 1 ∣ - \frac{1}{x - 1})

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

= 7 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - \frac{3}{x - 1}

Simplify

7 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - \frac{3}{x - 1} : 7 ln ∣ x - 1 ∣ - \frac{10}{x - 1}

= 7 ln ∣ x - 1 ∣ - \frac{10}{x - 1}

Add a constant to the solution

= 7 ln ∣ x - 1 ∣ - \frac{10}{x - 1} + C

\frac{e ^{2 s}}{s ^{3}}

\frac{\partial}{\partial x} (x y e^{s i n (π x y)})

x \to - 3 lim ((5 - x) \frac{4}{3})

\int \frac{5}{1 + 9 x ^{2}} d x

\int y^{2} cosh (7 y) d y

What is the integral of (7x+3)/(x^2-2x+1) ?
The integral of (7x+3)/(x^2-2x+1) is 7ln|x-1|-(10)/(x-1)+C