What is the integral of (x^2)/(x-10) ?

The integral of (x^2)/(x-10) is ((x-10)^2)/2+20(x-10)+100ln|x-10|+C

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integral of (x^2)/(x-10)

\int \frac{x ^{2}}{x - 10} d x

\frac{( x - 10 ) ^{2}}{2} + 20 (x - 10) + 100 ln ∣ x - 10 ∣ + C

Solution steps

\int \frac{x ^{2}}{x - 10} d x

Apply u-substitution

: \int \frac{( u + 10 ) ^{2}}{u} d u

= \int \frac{( u + 10 ) ^{2}}{u} d u

Expand

\frac{( u + 10 ) ^{2}}{u} : u + 20 + \frac{100}{u}

= \int u + 20 + \frac{100}{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 20 d u + \int \frac{100}{u} d u

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

\int 20 d u = 20 u

\int \frac{100}{u} d u = 100 ln ∣ u ∣

= \frac{u ^{2}}{2} + 20 u + 100 ln ∣ u ∣

Substitute back

u = x - 10

= \frac{( x - 10 ) ^{2}}{2} + 20 (x - 10) + 100 ln ∣ x - 10 ∣

Add a constant to the solution

= \frac{( x - 10 ) ^{2}}{2} + 20 (x - 10) + 100 ln ∣ x - 10 ∣ + C

\frac{d y}{d x} = (4 x + y)^{2}

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

\frac{\partial}{\partial x} (ln (10 x))

\frac{d}{d x} (\frac{x}{lo g _{e} ( x )})

\frac{d}{d x} (b e^{- x})

What is the integral of (x^2)/(x-10) ?
The integral of (x^2)/(x-10) is ((x-10)^2)/2+20(x-10)+100ln|x-10|+C