Indefinite IntegrationhardFree

Integral of (x⁷ᵐ+x²ᵐ+xᵐ)(2x⁶ᵐ+7xᵐ+14)^(1/m) | JEE

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

Question

For any natural number

m

and

x>0

\displaystyle\int (x^{7m}+x^{2m}+x^{m})\,(2x^{6m}+7x^{m}+14)^{1/m}\,dx

equals

\dfrac{(7x^{7m}+2x^{2m}+14x^{m})^{(m+1)/m}}{14(m+1)}+C

\dfrac{(2x^{7m}+14x^{2m}+7x^{m})^{(m+1)/m}}{14(m+1)}+C

\dfrac{(2x^{7m}+7x^{2m}+14x^{m})^{(m+1)/m}}{14(m+1)}+C

correct

\dfrac{(7x^{7m}+2x^{2m}+x^{m})^{(m+1)/m}}{14(m+1)}+C

Solution

Step 1: Take a factor

x

inside the

1/m

-power. Multiplying the base by

x^{m}

multiplies the power by

x

\big(x^{m}(2x^{6m}+7x^{m}+14)\big)^{1/m}=(2x^{7m}+7x^{2m}+14x^{m})^{1/m}=x\,(2x^{6m}+7x^{m}+14)^{1/m}

\Rightarrow (2x^{6m}+7x^{m}+14)^{1/m}=\dfrac{1}{x}\,(2x^{7m}+7x^{2m}+14x^{m})^{1/m}

Step 2: Multiply this into the outer factor, lowering each exponent by

1

I=\int (x^{7m-1}+x^{2m-1}+x^{m-1})\,(2x^{7m}+7x^{2m}+14x^{m})^{1/m}\,dx

Step 3: Substitute

t=2x^{7m}+7x^{2m}+14x^{m}

. Then

dt=\big(14m\,x^{7m-1}+14m\,x^{2m-1}+14m\,x^{m-1}\big)\,dx=14m\,(x^{7m-1}+x^{2m-1}+x^{m-1})\,dx

\Rightarrow (x^{7m-1}+x^{2m-1}+x^{m-1})\,dx=\dfrac{dt}{14m}

Step 4: The integral becomes

I=\dfrac{1}{14m}\int t^{1/m}\,dt=\dfrac{1}{14m}\cdot\dfrac{t^{\,1/m+1}}{\tfrac{1}{m}+1}

Step 5: Simplify the constant, using

\dfrac{1}{m}+1=\dfrac{m+1}{m}

, so

\dfrac{1}{14m}\cdot\dfrac{m}{m+1}=\dfrac{1}{14(m+1)}

I=\dfrac{t^{(m+1)/m}}{14(m+1)}+C=\dfrac{(2x^{7m}+7x^{2m}+14x^{m})^{(m+1)/m}}{14(m+1)}+C

Correct answer: (3)

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