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 mm and x>0x>0, ∫(x7m+x2m+xm) (2x6m+7xm+14)1/m dx\displaystyle\int (x^{7m}+x^{2m}+x^{m})\,(2x^{6m}+7x^{m}+14)^{1/m}\,dx equals
A(7x7m+2x2m+14xm)(m+1)/m14(m+1)+C\dfrac{(7x^{7m}+2x^{2m}+14x^{m})^{(m+1)/m}}{14(m+1)}+C
B(2x7m+14x2m+7xm)(m+1)/m14(m+1)+C\dfrac{(2x^{7m}+14x^{2m}+7x^{m})^{(m+1)/m}}{14(m+1)}+C
C(2x7m+7x2m+14xm)(m+1)/m14(m+1)+C\dfrac{(2x^{7m}+7x^{2m}+14x^{m})^{(m+1)/m}}{14(m+1)}+Ccorrect
D(7x7m+2x2m+xm)(m+1)/m14(m+1)+C\dfrac{(7x^{7m}+2x^{2m}+x^{m})^{(m+1)/m}}{14(m+1)}+C
Solution
Step 1: Take a factor xx inside the 1/m1/m-power. Multiplying the base by xmx^{m} multiplies the power by xx:
(xm(2x6m+7xm+14))1/m=(2x7m+7x2m+14xm)1/m=x (2x6m+7xm+14)1/m\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}
⇒(2x6m+7xm+14)1/m=1x (2x7m+7x2m+14xm)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 11:
I=∫(x7m−1+x2m−1+xm−1) (2x7m+7x2m+14xm)1/m dxI=\int (x^{7m-1}+x^{2m-1}+x^{m-1})\,(2x^{7m}+7x^{2m}+14x^{m})^{1/m}\,dx
 Step 3: Substitute t=2x7m+7x2m+14xmt=2x^{7m}+7x^{2m}+14x^{m}. Then
dt=(14m x7m−1+14m x2m−1+14m xm−1) dx=14m (x7m−1+x2m−1+xm−1) dxdt=\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
⇒(x7m−1+x2m−1+xm−1) dx=dt14m\Rightarrow (x^{7m-1}+x^{2m-1}+x^{m-1})\,dx=\dfrac{dt}{14m}
 Step 4: The integral becomes
I=114m∫t1/m dt=114m⋅t 1/m+11m+1I=\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 1m+1=m+1m\dfrac{1}{m}+1=\dfrac{m+1}{m}, so 114m⋅mm+1=114(m+1)\dfrac{1}{14m}\cdot\dfrac{m}{m+1}=\dfrac{1}{14(m+1)}:
I=t(m+1)/m14(m+1)+C=(2x7m+7x2m+14xm)(m+1)/m14(m+1)+CI=\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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