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

The integral of (x^2)/(x-5) is ((x-5)^2)/2+10(x-5)+25ln|x-5|+C

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

\int \frac{x ^{2}}{x - 5} d x

\frac{( x - 5 ) ^{2}}{2} + 10 (x - 5) + 25 ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{x ^{2}}{x - 5} d x

Apply u-substitution

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

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

Expand

\frac{( u + 5 ) ^{2}}{u} : u + 10 + \frac{25}{u}

= \int u + 10 + \frac{25}{u} 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 u d u + \int 10 d u + \int \frac{25}{u} d u

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

\int 10 d u = 10 u

\int \frac{25}{u} d u = 25 ln ∣ u ∣

= \frac{u ^{2}}{2} + 10 u + 25 ln ∣ u ∣

Substitute back

u = x - 5

= \frac{( x - 5 ) ^{2}}{2} + 10 (x - 5) + 25 ln ∣ x - 5 ∣

Add a constant to the solution

= \frac{( x - 5 ) ^{2}}{2} + 10 (x - 5) + 25 ln ∣ x - 5 ∣ + C

\frac{d y}{d x} = 0.2 \cdot y + 200

(x^{3} + y^{3}) d x - x y^{2} d y = 0

\frac{d}{d x} (\frac{x ^{2} ln ( x )}{1 + 2 x})

x - 7

- 10 x^{2} y + \frac{d y}{d x} = - 7 x^{2}

What is the integral of (x^2)/(x-5) ?
The integral of (x^2)/(x-5) is ((x-5)^2)/2+10(x-5)+25ln|x-5|+C