Integrals Involving Roots — Question 7

Question 7 – Solution

Step 1: Substitute $u=\sqrt[3]{x+1}$. Then $$u^3=x+1,\qquad x=u^3-1,\qquad dx=3u^2\,du.$$ Step 2: Rewrite and factor. $$\begin{align*} I&=\int\frac{3u^2\,du}{(u^3-1)u} =3\int\frac{u}{u^3-1}\,du,\\ u^3-1&=(u-1)(u^2+u+1). \end{align*}$$ Step 3: Use partial fractions. $$\frac{3u}{u^3-1}=\frac{A}{u-1}+\frac{Bu+C}{u^2+u+1}.$$ After clearing denominators, $$3u=A(u^2+u+1)+(Bu+C)(u-1).$$ Matching coefficients gives $$A+B=0,\qquad A-B+C=3,\qquad A-C=0,$$ so $A=1$, $B=-1$, and $C=1$. Therefore, $$\frac{3u}{u^3-1}=\frac1{u-1}+\frac{-u+1}{u^2+u+1}.$$ Step 4: Prepare the quadratic term. $$-u+1=-\frac12(2u+1)+\frac32.$$ Thus $$\begin{align*} I={}&\int\frac{du}{u-1} -\frac12\int\frac{2u+1}{u^2+u+1}\,du +\frac32\int\frac{du}{u^2+u+1}. \end{align*}$$ Complete the square: $$u^2+u+1=\left(u+\frac12\right)^2+\frac34.$$ Step 5: Integrate and return to $x$. $$\begin{align*} I={}&\ln|u-1|-\frac12\ln(u^2+u+1) +\sqrt3\arctan\left(\frac{2u+1}{\sqrt3}\right)+C, \end{align*}$$ where $u=\sqrt[3]{x+1}$. $$\boxed{\displaystyle \ln|u-1|-\frac12\ln(u^2+u+1) +\sqrt3\arctan\left(\frac{2u+1}{\sqrt3}\right)+C}$$