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

The integral of (x^2)/((x-y)) is ((-y+x)^2)/2+2y(-y+x)+y^2ln|-y+x|+C

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

integral

\frac{x ^{2}}{( x - y )}

\frac{( - y + x ) ^{2}}{2} + 2 y (- y + x) + y^{2} ln ∣ - y + x ∣ + C

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{( u + y ) ^{2}}{u} : u + 2 y + \frac{y ^{2}}{u}

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

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

\int 2 y d u = 2 y u

\int \frac{y ^{2}}{u} d u = y^{2} ln ∣ u ∣

= \frac{u ^{2}}{2} + 2 y u + y^{2} ln ∣ u ∣

Substitute back

u = - y + x

= \frac{( - y + x ) ^{2}}{2} + 2 y (- y + x) + y^{2} ln ∣ - y + x ∣

Add a constant to the solution

= \frac{( - y + x ) ^{2}}{2} + 2 y (- y + x) + y^{2} ln ∣ - y + x ∣ + C

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

x^{2} (a^{2} - x^{2})^{\frac{1}{2}}

\frac{( 2 x - 1 )}{( 2 x + 3 )}

\frac{cos ( 2 x )}{cos ( x )}

\frac{( x + 8 ) ^{2}}{x ^{2} + 7 x - 8}

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