Question 34

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Jambmaths question: 

Evaluate $\int{2{{(2x-3)}^{\tfrac{2}{3}}}dx}$

Option A: 

$\tfrac{3}{5}{{(2x-3)}^{\tfrac{5}{3}}}+k$

Option B: 

$\tfrac{6}{5}{{(2x-3)}^{\tfrac{5}{3}}}+k$

Option C: 

$2x-3+k$

Option D: 

$2(2x-3)+k$

Jamb Maths Solution: 

$\begin{align}  & I=\int{2{{(2x-3)}^{\tfrac{2}{3}}}dx} \\ & \text{Using change of variable method, }u=(2x-3) \\ & \frac{du}{dx}=2,\text{ }dx=\frac{du}{2} \\ & I=\not{2}\int{{{u}^{\tfrac{2}{3}}}\tfrac{du}{{\not{2}}}}=\int{{{u}^{\tfrac{2}{3}}}du} \\ & I=\frac{{{u}^{\tfrac{2}{3}+1}}}{\tfrac{2}{3}+1}=\frac{{{u}^{\tfrac{5}{3}}}}{\tfrac{5}{3}}+C \\ & I=\frac{3}{5}{{u}^{\tfrac{5}{3}}}+C=\frac{3}{5}{{(2x-3)}^{\tfrac{5}{3}}}+C \\\end{align}$

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