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

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

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

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

- \frac{x + 1}{x - 3} + ln ∣ x - 3 ∣ + C

Solution steps

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

Apply Integration By Parts

: - \frac{x + 1}{x - 3} - \int - \frac{1}{x - 3} d x

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

\int - \frac{1}{x - 3} d x = - ln ∣ x - 3 ∣

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

Simplify

= - \frac{x + 1}{x - 3} + ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{x + 1}{x - 3} + ln ∣ x - 3 ∣ + C

\int_{t}^{1} \frac{1}{x ^{2}} d x

x \to 8 lim (cos (\frac{π x}{3}))

y = \frac{x ^{4}}{f ( x )}

f (x) = 3 x^{2} - 4 x

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 4, y^{^{'}} (0) = - 2

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