Indefinite IntegrationmediumFree

Integral of 1/(√x − ∛x) | JEE

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

Question

\displaystyle\int \dfrac{dx}{\sqrt{x}-\sqrt[3]{x}}

equals

2\sqrt{x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\ln\big|\sqrt[6]{x}-1\big|+C

correct

2\sqrt{x}+3\sqrt[3]{x}-6\sqrt[6]{x}+6\ln\big|\sqrt[6]{x}-1\big|+C

3\sqrt{x}+2\sqrt[3]{x}+6\sqrt[6]{x}+6\ln\big|\sqrt[6]{x}-1\big|+C

3\sqrt{x}+2\sqrt[3]{x}-6\sqrt[6]{x}-\ln\big|\sqrt[6]{x}-1\big|+C

Solution

Step 1: Substitute

x=t^{6}

(the LCM of the root orders

2

and

3

). Then

dx=6t^{5}\,dt

\sqrt{x}=t^{3}

\sqrt[3]{x}=t^{2}

. Step 2: Rewrite the integral:

I=\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

t^{3}

(t-1)

\dfrac{t^{3}}{t-1}=t^{2}+t+1+\dfrac{1}{t-1}

Step 4: Integrate term by term:

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=x^{1/6}

(so

t^{3}=\sqrt{x},\ t^{2}=\sqrt[3]{x},\ t=\sqrt[6]{x}

I=2\sqrt{x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\ln\big|\sqrt[6]{x}-1\big|+C

Correct answer: (1)

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