The integral \int ( 1 + x - 1 \over x )e^x + 1 \over xdx is equal to

Indefinite Integrals PYQ — JEE Main 2014 · Shift 0 · Q30 with Solution

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Correct answer: D

We need to evaluate $I=\int \left(1+x-\frac1x\right)e^{x+\frac1x}\,dx.$
Look at the exponent: $u=x+\frac1x.$ Then $\frac{du}{dx}=1-\frac1{x^2}.$ This does not directly match the factor $\left(1+x-\frac1x\right)$ , so we try to rewrite the integrand in a useful way.
A standard trick with expressions of the form $e^{x+1/x}$ is to differentiate something like $x e^{x+1/x}$ :

= e^{x+\frac1x}+x e^{x+\frac1x}\left(1-\frac1{x^2}\right).$$ Now simplify: $$\frac{d}{dx}\left(xe^{x+\frac1x}\right) = e^{x+\frac1x}\left[1+x\left(1-\frac1{x^2}\right)\right] = e^{x+\frac1x}\left(1+x-\frac1x\right).$$ 4. This exactly matches the integrand. Therefore, $$I=xe^{x+\frac1x}+C.$$ 5. Hence the integral is $$\boxed{xe^{x+\frac1x}+C}.$$ 6. Comparing with the stored correct answer: the derived result corresponds to option $D$ (as given in the prompt). So the stored answer is consistent.

View written solutionFree

Correct answer: D

We need to evaluate $I=\int \left(1+x-\frac1x\right)e^{x+\frac1x}\,dx.$
Look at the exponent: $u=x+\frac1x.$ Then $\frac{du}{dx}=1-\frac1{x^2}.$ This does not directly match the factor $\left(1+x-\frac1x\right)$ , so we try to rewrite the integrand in a useful way.
A standard trick with expressions of the form $e^{x+1/x}$ is to differentiate something like $x e^{x+1/x}$ :

= e^{x+\frac1x}+x e^{x+\frac1x}\left(1-\frac1{x^2}\right).$$ Now simplify: $$\frac{d}{dx}\left(xe^{x+\frac1x}\right) = e^{x+\frac1x}\left[1+x\left(1-\frac1{x^2}\right)\right] = e^{x+\frac1x}\left(1+x-\frac1x\right).$$ 4. This exactly matches the integrand. Therefore, $$I=xe^{x+\frac1x}+C.$$ 5. Hence the integral is $$\boxed{xe^{x+\frac1x}+C}.$$ 6. Comparing with the stored correct answer: the derived result corresponds to option $D$ (as given in the prompt). So the stored answer is consistent.