Evaluate the integral: I = integral (x^4)/((x+1)³) dx

Evaluate the integral: $$ I = \int \frac{x^4}{(x+1)^3}\,dx. $$

Step 1 · Polynomial long division

Expanding $(x+1)^3=x^3+3x^2+3x+1$ and dividing:

$$ \frac{x^4}{(x+1)^3} = x-3 + \frac{6x^2+8x+3}{(x+1)^3}. $$

Step 2 · Integrate the polynomial part

$$ \int (x-3)\,dx = \frac{x^2}{2} - 3x. $$

Step 3 · Substitute for the remaining fraction

Let $u=x+1$, $x=u-1$, $du=dx$:

$$ 6x^2+8x+3 = 6(u-1)^2+8(u-1)+3 = 6u^2-4u+1. $$

$$ \int \frac{6x^2+8x+3}{(x+1)^3}\,dx = \int \left(\frac{6}{u} - \frac{4}{u^2} + \frac{1}{u^3}\right) du = 6\ln|u| + \frac4u - \frac{1}{2u^2}. $$

Step 4 · Combine and substitute back

$$ I = \frac{x^2}{2} - 3x + 6\ln|x+1| + \frac{4}{x+1} - \frac{1}{2(x+1)^2} + C. $$

Summary Table

Additional Notes

• The domain excludes $x=-1$, where the denominator vanishes.
• Because the remainder $6x^2+8x+3$ has lower degree than $(x+1)^3$, no further division is needed after the substitution — expressing it in powers of $u=x+1$ directly splits it into three simple power-rule terms.
• The result can be checked by differentiating $\tfrac{x^2}2-3x+6\ln|x+1|+\tfrac4{x+1}-\tfrac{1}{2(x+1)^2}$: combining all terms over $(x+1)^3$ and simplifying the numerator reduces exactly to $x^4$, confirmed by direct polynomial expansion.