Evaluate the integral: int {frac{{{e^{sqrt x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}} \left( {x + \sqrt x } \right)dx$. Substitute $\sqrt x = t$,then $\frac{1}{{2\sqrt x }}dx = dt$

Vedclass · Accepted Answer

$2{e^{\sqrt x }}\left[ {x - \sqrt x + 1} \right] + C$

Vedclass · Answer

(A) Let $I = \int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}} \left( {x + \sqrt x } \right)dx$. Substitute $\sqrt x = t$,then $\frac{1}{{2\sqrt x }}dx = dt$,which implies $\frac{{dx}}{{\sqrt x }} = 2dt$. Substituting these into the integral,we get: $I = \int {{e^t}} (t^2 + t) (2dt) = 2 \int {{e^t}} (t^2 + t) dt$. We know that $\int {{e^t}} (f(t) + f'(t)) dt = {e^t}f(t) + C$. Here,let $f(t) = t^2$. Then $f'(t) = 2t$. This does not match directly. Let's rewrite the integral: $I = 2 \int {{e^t}} (t^2 + t) dt$. Using integration by parts or observing the form: $I = 2 [\int {{e^t}} t^2 dt + \int {{e^t}} t dt]$. Alternatively,consider $f(t) = t^2 - t + 1$. Then $f'(t) = 2t - 1$. Actually,let's use the substitution $I = 2 \int {{e^t}} (t^2 + t) dt$. Using $\int {{e^t}} (t^2 + 2t - t) dt = \int {{e^t}} (t^2 + 2t) dt - \int {{e^t}} t dt$. This is $\int {{e^t}} (t^2 + 2t) dt - [t{e^t} - \int {{e^t}} dt] = {e^t}{t^2} - t{e^t} + {e^t} + C = {e^t}(t^2 - t + 1) + C$. Multiplying by the constant $2$ from the substitution: $I = 2{e^t}(t^2 - t + 1) + C$. Substituting $t = \sqrt x$: $I = 2{e^{\sqrt x }}(x - \sqrt x + 1) + C$.

Evaluate the integral: $\int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}} \left( {x + \sqrt x } \right)dx$

Similar Questions

If $\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c$,then find $f(x)$.

$ \int e^{\sin x} \cdot \left(\frac{\sin x+1}{\sec x}\right) d x $ is equal to

$\int e^x \frac{x^2+1}{(x+1)^2} d x$ is equal to

$\int e^x \left( \frac{\sec^2 x + \tan x - \cot x}{\sin x} \right) dx =$

$\int e^x \left( \frac{2 + \sin 2x}{1 + \cos 2x} \right) dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module