What is the integral of (x^4)/((1-x)^3) ?

The integral of (x^4)/((1-x)^3) is (x^4)/(2(1-x)^2)-2((x^3)/(1-x)+3(1/2 x^2+x+ln|-x+1|+1/2-2))+C

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integral of (x^4)/((1-x)^3)

\int \frac{x ^{4}}{( 1 - x ) ^{3}} d x

\frac{x ^{4}}{2 ( 1 - x ) ^{2}} - 2 (\frac{x ^{3}}{1 - x} + 3 (\frac{1}{2} x^{2} + x + ln ∣ - x + 1 ∣ + \frac{1}{2} - 2)) + C

Solution steps

\int \frac{x ^{4}}{( 1 - x ) ^{3}} d x

Apply Integration By Parts

: \frac{x ^{4}}{2 ( 1 - x ) ^{2}} - \int \frac{2 x ^{3}}{( 1 - x ) ^{2}} d x

= \frac{x ^{4}}{2 ( 1 - x ) ^{2}} - \int \frac{2 x ^{3}}{( 1 - x ) ^{2}} d x

\int \frac{2 x ^{3}}{( 1 - x ) ^{2}} d x = 2 (\frac{x ^{3}}{1 - x} + 3 (\frac{1}{2} x^{2} + x + ln ∣ - x + 1 ∣ + \frac{1}{2} - 2))

= \frac{x ^{4}}{2 ( 1 - x ) ^{2}} - 2 (\frac{x ^{3}}{1 - x} + 3 (\frac{1}{2} x^{2} + x + ln ∣ - x + 1 ∣ + \frac{1}{2} - 2))

Add a constant to the solution

= \frac{x ^{4}}{2 ( 1 - x ) ^{2}} - 2 (\frac{x ^{3}}{1 - x} + 3 (\frac{1}{2} x^{2} + x + ln ∣ - x + 1 ∣ + \frac{1}{2} - 2)) + C

f (x) = (x^{4} + 3 x)^{- 1}

\int sin^{2} (θ) tan (θ) d θ

\frac{\partial}{\partial x} (e^{x} + y + y^{2} + 6)

x \to 0 lim (ln (1 - x))

\int (cos^{4} (x)) (sin (x)) d x

What is the integral of (x^4)/((1-x)^3) ?
The integral of (x^4)/((1-x)^3) is (x^4)/(2(1-x)^2)-2((x^3)/(1-x)+3(1/2 x^2+x+ln|-x+1|+1/2-2))+C