What is the integral of (x^2)/(x-9) ?

The integral of (x^2)/(x-9) is ((x-9)^2)/2+18(x-9)+81ln|x-9|+C

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integral of (x^2)/(x-9)

\int \frac{x ^{2}}{x - 9} d x

\frac{( x - 9 ) ^{2}}{2} + 18 (x - 9) + 81 ln ∣ x - 9 ∣ + C

Solution steps

\int \frac{x ^{2}}{x - 9} d x

Apply u-substitution

: \int \frac{( u + 9 ) ^{2}}{u} d u

= \int \frac{( u + 9 ) ^{2}}{u} d u

Expand

\frac{( u + 9 ) ^{2}}{u} : u + 18 + \frac{81}{u}

= \int u + 18 + \frac{81}{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 18 d u + \int \frac{81}{u} d u

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

\int 18 d u = 18 u

\int \frac{81}{u} d u = 81 ln ∣ u ∣

= \frac{u ^{2}}{2} + 18 u + 81 ln ∣ u ∣

Substitute back

u = x - 9

= \frac{( x - 9 ) ^{2}}{2} + 18 (x - 9) + 81 ln ∣ x - 9 ∣

Add a constant to the solution

= \frac{( x - 9 ) ^{2}}{2} + 18 (x - 9) + 81 ln ∣ x - 9 ∣ + C

2 - x^{2}, x^{2}

\frac{\partial}{\partial x} (x^{2} + 2 y^{2} + 9 x z - 12)

\int tan^{2} (x) sec^{2} (x) d x

x = 0, y = 15 e^{- x^{2}}, [0, 1]

What is the integral of (x^2)/(x-9) ?
The integral of (x^2)/(x-9) is ((x-9)^2)/2+18(x-9)+81ln|x-9|+C