The value of the integral \int_-\log _e 2^\log _e 2 e^x(\log _e(e^x+\sqrt1+e^2 x)) d x is equal to :

\log _e(\frac\sqrt2(2+\sqrt5)^2\sqrt1+\sqrt5)-\frac\sqrt52

\log _e(\frac(2+\sqrt5)^2\sqrt1+\sqrt5)+\frac\sqrt52

\log _e(\frac2(2+\sqrt5)\sqrt1+\sqrt5)-\frac\sqrt52

\log _e(\frac\sqrt2(3-\sqrt5)^2\sqrt1+\sqrt5)+\frac\sqrt52

Definite Integration PYQ — JEE Main 2023 · 11 Apr · Shift 1 · Q30 with Solution

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Correct answer: B

\left(e^x+\sqrt{1+e^{2x}}\right)\,dx.$$ Notice that $$\ln\left(t+\sqrt{1+t^2}\right)=\sinh^{-1}(t).$$ So with $t=e^x$, the integrand suggests a substitution. 2. Put $$t=e^x \implies dt=e^x dx.$$ The limits change as: - when $x=-\ln 2$, $t=e^{-\ln 2}=\frac12$, - when $x=\ln 2$, $t=e^{\ln 2}=2$. Therefore, $$I=\int_{1/2}^{2}\ln\left(t+\sqrt{1+t^2}\right)\,dt.$$ 3. Use the standard antiderivative $$\int \ln\left(t+\sqrt{1+t^2}\right)\,dt = t\ln\left(t+\sqrt{1+t^2}\right)-\sqrt{1+t^2}+C,$$ because $\ln(t+\sqrt{1+t^2})=\sinh^{-1}t$ and $$\int \sinh^{-1}t\,dt=t\sinh^{-1}t-\sqrt{1+t^2}+C.$$ Hence, $$I=\left[t\ln\left(t+\sqrt{1+t^2}\right)-\sqrt{1+t^2}\right]_{1/2}^{2}.$$ 4. Evaluate at the upper limit $t=2$: $$\sqrt{1+2^2}=\sqrt5,$$ so $$F(2)=2\ln(2+\sqrt5)-\sqrt5.$$ At the lower limit $t=\frac12$: $$\sqrt{1+\left(\frac12\right)^2}=\frac{\sqrt5}{2},$$ thus $$F\left(\frac12\right)=\frac12\ln\left(\frac12+\frac{\sqrt5}{2}\right)-\frac{\sqrt5}{2} =\frac12\ln\left(\frac{1+\sqrt5}{2}\right)-\frac{\sqrt5}{2}.$$ Therefore, \begin{align*} I&=\left(2\ln(2+\sqrt5)-\sqrt5\right)-\left(\frac12\ln\left(\frac{1+\sqrt5}{2}\right)-\frac{\sqrt5}{2}\right)\\ &=2\ln(2+\sqrt5)-\frac12\ln\left(\frac{1+\sqrt5}{2}\right)-\frac{\sqrt5}{2}. \end{align*} 5. Simplify the logarithmic part: \begin{align*} 2\ln(2+\sqrt5)-\frac12\ln\left(\frac{1+\sqrt5}{2}\right) &=\ln(2+\sqrt5)^2-\ln\left(\frac{1+\sqrt5}{2}\right)^{1/2}\\ &=\ln\left(\frac{(2+\sqrt5)^2}{\sqrt{(1+\sqrt5)/2}}\right)\\ &=\ln\left(\frac{\sqrt2(2+\sqrt5)^2}{\sqrt{1+\sqrt5}}\right). \end{align*} So, $$I=\ln\left(\frac{\sqrt2(2+\sqrt5)^2}{\sqrt{1+\sqrt5}}\right)-\frac{\sqrt5}{2}.$$ 6. Comparing with the options, this is exactly **Option B**.

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View written solutionFree

Correct answer: B

\left(e^x+\sqrt{1+e^{2x}}\right)\,dx.$$ Notice that $$\ln\left(t+\sqrt{1+t^2}\right)=\sinh^{-1}(t).$$ So with $t=e^x$, the integrand suggests a substitution. 2. Put $$t=e^x \implies dt=e^x dx.$$ The limits change as: - when $x=-\ln 2$, $t=e^{-\ln 2}=\frac12$, - when $x=\ln 2$, $t=e^{\ln 2}=2$. Therefore, $$I=\int_{1/2}^{2}\ln\left(t+\sqrt{1+t^2}\right)\,dt.$$ 3. Use the standard antiderivative $$\int \ln\left(t+\sqrt{1+t^2}\right)\,dt = t\ln\left(t+\sqrt{1+t^2}\right)-\sqrt{1+t^2}+C,$$ because $\ln(t+\sqrt{1+t^2})=\sinh^{-1}t$ and $$\int \sinh^{-1}t\,dt=t\sinh^{-1}t-\sqrt{1+t^2}+C.$$ Hence, $$I=\left[t\ln\left(t+\sqrt{1+t^2}\right)-\sqrt{1+t^2}\right]_{1/2}^{2}.$$ 4. Evaluate at the upper limit $t=2$: $$\sqrt{1+2^2}=\sqrt5,$$ so $$F(2)=2\ln(2+\sqrt5)-\sqrt5.$$ At the lower limit $t=\frac12$: $$\sqrt{1+\left(\frac12\right)^2}=\frac{\sqrt5}{2},$$ thus $$F\left(\frac12\right)=\frac12\ln\left(\frac12+\frac{\sqrt5}{2}\right)-\frac{\sqrt5}{2} =\frac12\ln\left(\frac{1+\sqrt5}{2}\right)-\frac{\sqrt5}{2}.$$ Therefore, \begin{align*} I&=\left(2\ln(2+\sqrt5)-\sqrt5\right)-\left(\frac12\ln\left(\frac{1+\sqrt5}{2}\right)-\frac{\sqrt5}{2}\right)\\ &=2\ln(2+\sqrt5)-\frac12\ln\left(\frac{1+\sqrt5}{2}\right)-\frac{\sqrt5}{2}. \end{align*} 5. Simplify the logarithmic part: \begin{align*} 2\ln(2+\sqrt5)-\frac12\ln\left(\frac{1+\sqrt5}{2}\right) &=\ln(2+\sqrt5)^2-\ln\left(\frac{1+\sqrt5}{2}\right)^{1/2}\\ &=\ln\left(\frac{(2+\sqrt5)^2}{\sqrt{(1+\sqrt5)/2}}\right)\\ &=\ln\left(\frac{\sqrt2(2+\sqrt5)^2}{\sqrt{1+\sqrt5}}\right). \end{align*} So, $$I=\ln\left(\frac{\sqrt2(2+\sqrt5)^2}{\sqrt{1+\sqrt5}}\right)-\frac{\sqrt5}{2}.$$ 6. Comparing with the options, this is exactly **Option B**.

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