What is the integral of (5x^2)/((x-7)^3) ?

The integral of (5x^2)/((x-7)^3) is 5(ln|x-7|-(14)/(x-7)-(49)/(2(x-7)^2))+C

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integral of (5x^2)/((x-7)^3)

\int \frac{5 x ^{2}}{( x - 7 ) ^{3}} d x

5 (ln ∣ x - 7 ∣ - \frac{14}{x - 7} - \frac{49}{2 ( x - 7 ) ^{2}}) + C

Solution steps

\int \frac{5 x ^{2}}{( x - 7 ) ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int \frac{x ^{2}}{( x - 7 ) ^{3}} d x

Apply u-substitution

: \int \frac{( u + 7 ) ^{2}}{u ^{3}} d u

= 5 \cdot \int \frac{( u + 7 ) ^{2}}{u ^{3}} d u

Expand

\frac{( u + 7 ) ^{2}}{u ^{3}} : \frac{1}{u} + \frac{14}{u ^{2}} + \frac{49}{u ^{3}}

= 5 \cdot \int \frac{1}{u} + \frac{14}{u ^{2}} + \frac{49}{u ^{3}} 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

= 5 (\int \frac{1}{u} d u + \int \frac{14}{u ^{2}} d u + \int \frac{49}{u ^{3}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{14}{u ^{2}} d u = - \frac{14}{u}

\int \frac{49}{u ^{3}} d u = - \frac{49}{2 u ^{2}}

= 5 (ln ∣ u ∣ - \frac{14}{u} - \frac{49}{2 u ^{2}})

Substitute back

u = x - 7

= 5 (ln ∣ x - 7 ∣ - \frac{14}{x - 7} - \frac{49}{2 ( x - 7 ) ^{2}})

Add a constant to the solution

= 5 (ln ∣ x - 7 ∣ - \frac{14}{x - 7} - \frac{49}{2 ( x - 7 ) ^{2}}) + C

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What is the integral of (5x^2)/((x-7)^3) ?
The integral of (5x^2)/((x-7)^3) is 5(ln|x-7|-(14)/(x-7)-(49)/(2(x-7)^2))+C