Question

If $\int{\dfrac{x+1}{\sqrt{2x-1}}}dx=f\left( x \right)\sqrt{2x-1}+C$ where C is a constant of integration then f (x) is equal to:

& A.\dfrac{1}{3}\left( x+4 \right) \\\ & B.\dfrac{1}{3}\left( x+1 \right) \\\ & C.\dfrac{2}{3}\left( x+2 \right) \\\ & D.\dfrac{2}{3}\left( x-4 \right) \\\ \end{aligned}$$

Answer 1

By looking over the question, we found no limits of integration and constant of integration is present, hence it is indefinite integration, in which we have to evaluate left hand side expression and then compare it to RHS. In most of the cases in integration, whenever we see some variable present in under root expression (like here $\sqrt{2x-1}$) then we have to put that expression to some other variable. This process is known as a method of substitution. After substituting the value, we get the integration in more simple and easy computable form.

Complete step by step answer:
We have been given that,
$\int{\dfrac{x+1}{\sqrt{2x-1}}}dx=f\left( x \right)\sqrt{2x-1}+C\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$
Now, as we told earlier, put $\sqrt{2x-1}=t$
By squaring both sides, we get:
$2x-1={{t}^{2}}$
Now, differentiate both the sides, we have:

& 2dx=2tdt \\\ & \text{Cancelling 2 from both sides,} \\\ & dx=tdt\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\\ \end{aligned}$$ Now, we have: $$\begin{aligned} & 2x-1={{t}^{2}} \\\ & \therefore x=\left( \dfrac{1+{{t}^{2}}}{2} \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)} \\\ \end{aligned}$$ Now, from LHS part of the equation (i) we have: $$\begin{aligned} & \int{\dfrac{x+1}{\sqrt{2x-1}}}dx \\\ & \Rightarrow \int{\dfrac{\left( \dfrac{1+{{t}^{2}}}{2} \right)+1}{t}}\times tdt\text{ }\left( \text{from equation }\left( \text{ii} \right)\text{ and }\left( \text{iii} \right) \right) \\\ & \Rightarrow \int{\left( \dfrac{1+{{t}^{2}}+2}{2t} \right)}tdt \\\ & \Rightarrow \dfrac{1}{2}\int{\left( {{t}^{2}}+3 \right)dt} \\\ & \Rightarrow \dfrac{1}{2}\left( \dfrac{{{t}^{3}}}{3}+3t \right)+C \\\ & \Rightarrow \dfrac{1}{2}\times t\left( \dfrac{{{t}^{2}}}{3}+3 \right)+C \\\ & \Rightarrow \dfrac{1}{2}\times \sqrt{2x-1}\left( \dfrac{2x-1}{3}+3 \right)+C \\\ & \Rightarrow \dfrac{1}{2}\times \sqrt{2x-1}\left( \dfrac{2x+8}{3} \right)+C \\\ \end{aligned}$$ Now, by comparing with the RHS of equation (i) we have: $$f\left( x \right)=\left( \dfrac{2x+8}{3\times 2} \right)=\left( \dfrac{x+4}{3} \right)$$ **So, the correct answer is “Option A”.** **Note:** Sometimes, when students do the substitution method, when they convert the given integration to new integration form, they forget to change the (dx) part of the integration, which leads to wrong answers. We have to change each and every term with the newly assigned variable. These mistakes will be more blunder when we do definite integration, where we have to change the limits also.