Indefinite IntegrationmediumFree

Integral of 1/(√x − ∛x) | JEE

JEE Maths question with a full step-by-step solution.

Question
∫dxx−x3\displaystyle\int \dfrac{dx}{\sqrt{x}-\sqrt[3]{x}} equals
A2x+3x3+6x6+6ln⁡∣x6−1∣+C2\sqrt{x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\ln\big|\sqrt[6]{x}-1\big|+Ccorrect
B2x+3x3−6x6+6ln⁡∣x6−1∣+C2\sqrt{x}+3\sqrt[3]{x}-6\sqrt[6]{x}+6\ln\big|\sqrt[6]{x}-1\big|+C
C3x+2x3+6x6+6ln⁡∣x6−1∣+C3\sqrt{x}+2\sqrt[3]{x}+6\sqrt[6]{x}+6\ln\big|\sqrt[6]{x}-1\big|+C
D3x+2x3−6x6−ln⁡∣x6−1∣+C3\sqrt{x}+2\sqrt[3]{x}-6\sqrt[6]{x}-\ln\big|\sqrt[6]{x}-1\big|+C
Solution
Step 1: Substitute x=t6x=t^{6} (the LCM of the root orders 22 and 33). Then dx=6t5 dtdx=6t^{5}\,dt, x=t3\sqrt{x}=t^{3}, x3=t2\sqrt[3]{x}=t^{2}. Step 2: Rewrite the integral:
I=∫6t5t3−t2 dt=∫6t5t2(t−1) dt=∫6t3t−1 dtI=\int\dfrac{6t^{5}}{t^{3}-t^{2}}\,dt=\int\dfrac{6t^{5}}{t^{2}(t-1)}\,dt=\int\dfrac{6t^{3}}{t-1}\,dt
 Step 3: Divide t3t^{3} by (t−1)(t-1):
t3t−1=t2+t+1+1t−1\dfrac{t^{3}}{t-1}=t^{2}+t+1+\dfrac{1}{t-1}
 Step 4: Integrate term by term:
I=6∫ ⁣(t2+t+1+1t−1)dt=6 ⁣(t33+t22+t+ln⁡∣t−1∣)=2t3+3t2+6t+6ln⁡∣t−1∣I=6\int\!\left(t^{2}+t+1+\dfrac{1}{t-1}\right)dt=6\!\left(\dfrac{t^{3}}{3}+\dfrac{t^{2}}{2}+t+\ln|t-1|\right)=2t^{3}+3t^{2}+6t+6\ln|t-1|
 Step 5: Back-substitute t=x1/6t=x^{1/6} (so t3=x, t2=x3, t=x6t^{3}=\sqrt{x},\ t^{2}=\sqrt[3]{x},\ t=\sqrt[6]{x}):
I=2x+3x3+6x6+6ln⁡∣x6−1∣+CI=2\sqrt{x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\ln\big|\sqrt[6]{x}-1\big|+C
 Correct answer: (1)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Indefinite Integration · medium
If ∫ex 1−x6−6x2(1+x3)2 dx=ex 1−xp1+xq+c\displaystyle\int e^{x}\,\dfrac{1-x^{6}-6x^{2}}{(1+x^{3})^{2}}\,dx=e^{x}\,\dfrac{1-x^{p}}{1+x^{q}}+c, then p+qp+q equals
Indefinite Integration · easy
If ∫x−sin⁡xcos⁡xx2cos⁡2x dx=f(x)+c\displaystyle\int \dfrac{x-\sin x\cos x}{x^{2}\cos^{2}x}\,dx=f(x)+c, then lim⁡x→0f(x)\displaystyle\lim_{x\to 0}f(x) equals
Indefinite Integration · medium
If ∫x2e−2x dx=e−2x(ax2+bx+c)+d\displaystyle\int x^{2}e^{-2x}\,dx=e^{-2x}(ax^{2}+bx+c)+d, then
Indefinite Integration · medium
∫ex(2tan⁡x1+tan⁡x+cot⁡2 ⁣(x+π4))dx\displaystyle\int e^{x}\left(\dfrac{2\tan x}{1+\tan x}+\cot^{2}\!\left(x+\dfrac{\pi}{4}\right)\right)dx equals

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath