What is the integral of (x^2)/((x-3)(x-4)^2) ?

The integral of (x^2)/((x-3)(x-4)^2) is 9ln|x-3|-8ln|x-4|-(16)/(x-4)+C

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integral of (x^2)/((x-3)(x-4)^2)

\int \frac{x ^{2}}{( x - 3 ) ( x - 4 ) ^{2}} d x

9 ln ∣ x - 3 ∣ - 8 ln ∣ x - 4 ∣ - \frac{16}{x - 4} + C

Solution steps

\int \frac{x ^{2}}{( x - 3 ) ( x - 4 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 3 ) ( x - 4 ) ^{2}} : \frac{9}{x - 3} - \frac{8}{x - 4} + \frac{16}{( x - 4 ) ^{2}}

= \int \frac{9}{x - 3} - \frac{8}{x - 4} + \frac{16}{( x - 4 ) ^{2}} d x

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{9}{x - 3} d x - \int \frac{8}{x - 4} d x + \int \frac{16}{( x - 4 ) ^{2}} d x

\int \frac{9}{x - 3} d x = 9 ln ∣ x - 3 ∣

\int \frac{8}{x - 4} d x = 8 ln ∣ x - 4 ∣

\int \frac{16}{( x - 4 ) ^{2}} d x = - \frac{16}{x - 4}

= 9 ln ∣ x - 3 ∣ - 8 ln ∣ x - 4 ∣ - \frac{16}{x - 4}

Add a constant to the solution

= 9 ln ∣ x - 3 ∣ - 8 ln ∣ x - 4 ∣ - \frac{16}{x - 4} + C

x \to 0 lim (\frac{6 ^{x} - 8 ^{x}}{x})

y^{^{'}} = \frac{1}{y}

\frac{1}{2 \cdot 5} y + y^{^{'}} = \frac{t}{2}, y (30) = 10

(- 10, 1), (0, - 4)

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

What is the integral of (x^2)/((x-3)(x-4)^2) ?
The integral of (x^2)/((x-3)(x-4)^2) is 9ln|x-3|-8ln|x-4|-(16)/(x-4)+C