Question 42

Jambmaths question: 

Evaluate $\int{{{(2x+3)}^{\tfrac{1}{2}}}dx}$

Option A: 

$\tfrac{1}{3}{{(2x+3)}^{\tfrac{1}{2}}}+k$

Option B: 

$\tfrac{1}{3}{{(2x+3)}^{\tfrac{3}{2}}}+k$

Option C: 

$\tfrac{1}{12}{{(2x+3)}^{\tfrac{3}{4}}}+k$

Option D: 

$\tfrac{1}{12}{{(2x+3)}^{6}}+k$

Jamb Maths Solution: 

$\begin{align}  & y=\int{{{(2x+3)}^{\tfrac{1}{2}}}dx} \\ & \text{Using change of vairable method} \\ & \text{Let }u=2x+3,\text{  }\frac{du}{dx}=2,\text{ }dx=\frac{du}{2} \\ & y=\int{{{u}^{\tfrac{1}{2}}}\frac{du}{2}}=\frac{1}{2}\int{{{u}^{\tfrac{1}{2}}}}du=\frac{1}{2}\left[ \frac{{{u}^{\tfrac{1}{2}+1}}}{\tfrac{1}{2}+1} \right]+K \\ & y=\frac{1}{2}\left[ \frac{{{u}^{\tfrac{3}{2}}}}{\tfrac{3}{2}} \right]+K=\frac{1}{3}{{u}^{\tfrac{3}{2}}}+K \\ & y=\frac{1}{3}{{(2x+3)}^{\tfrac{3}{2}}}+K \\\end{align}$

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