Let f( x ) = \int \sqrt x \over ( 1 + x )^2dx( x \ge 0 ) . Then f(3) – f(1) is eqaul to :

\pi \over 12 + 1 \over 2 - \sqrt 3 \over 4

- \pi \over 12 + 1 \over 2 + \sqrt 3 \over 4

- \pi \over 6 + 1 \over 2 + \sqrt 3 \over 4

\pi \over 6 + 1 \over 2 - \sqrt 3 \over 4

Indefinite Integrals PYQ — JEE Main 2020 · 4 Sep · Shift 1 · Q34 with Solution

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Correct answer: B

Since $f(x)=\int \frac{\sqrt{x}}{(1+x)^2}\,dx,$ this is an indefinite integral, so $f(3)-f(1)$ can be evaluated as the corresponding definite integral: $f(3)-f(1)=\int_1^3 \frac{\sqrt{x}}{(1+x)^2}\,dx.$
Use the substitution $x=t^2, \quad dx=2t\,dt, \quad \sqrt{x}=t.$ Then when $x=1$ , $t=1$ , and when $x=3$ , $t=\sqrt{3}$ .

So,

=\int_1^{\sqrt{3}} \frac{t}{(1+t^2)^2}(2t\,dt) =\int_1^{\sqrt{3}} \frac{2t^2}{(1+t^2)^2}\,dt.$$ 3. Simplify the integrand: $$\frac{2t^2}{(1+t^2)^2} =\frac{2(1+t^2)-2}{(1+t^2)^2} =\frac{2}{1+t^2}-\frac{2}{(1+t^2)^2}.$$ Hence, $$f(3)-f(1)=\int_1^{\sqrt{3}} \frac{2}{1+t^2}\,dt-\int_1^{\sqrt{3}} \frac{2}{(1+t^2)^2}\,dt.$$ 4. Now use the standard result $$\int \frac{1}{(1+t^2)^2}\,dt=\frac{1}{2}\tan^{-1}t+\frac{t}{2(1+t^2)}+C.$$ Therefore, $$\int \frac{2}{(1+t^2)^2}\,dt=\tan^{-1}t+\frac{t}{1+t^2}+C.$$ So an antiderivative of $$\frac{2t^2}{(1+t^2)^2}$$ is $$2\tan^{-1}t-\left(\tan^{-1}t+\frac{t}{1+t^2}\right) =\tan^{-1}t-\frac{t}{1+t^2}.$$ Thus, $$f(3)-f(1)=\left[\tan^{-1}t-\frac{t}{1+t^2}\right]_1^{\sqrt{3}}.$$ 5. Evaluate at the limits: - At $t=\sqrt{3}$: $$\tan^{-1}(\sqrt{3})=\frac{\pi}{3}, \qquad \frac{\sqrt{3}}{1+3}=\frac{\sqrt{3}}{4}.$$ So the value is $$\frac{\pi}{3}-\frac{\sqrt{3}}{4}.$$ - At $t=1$: $$\tan^{-1}(1)=\frac{\pi}{4}, \qquad \frac{1}{1+1}=\frac{1}{2}.$$ So the value is $$\frac{\pi}{4}-\frac{1}{2}.$$ Therefore, $$f(3)-f(1)=\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)-\left(\frac{\pi}{4}-\frac{1}{2}\right).$$ 6. Simplify: $$f(3)-f(1)=\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}.$$ So the correct option is: $$\boxed{\text{B}}$$ with value $$\boxed{\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}}.$$ 7. Comparison with stored correct answer: - Stored correct answer: $\text{B}$ - Derived answer: $\text{B}$ They match.

View written solutionFree

Correct answer: B

Since $f(x)=\int \frac{\sqrt{x}}{(1+x)^2}\,dx,$ this is an indefinite integral, so $f(3)-f(1)$ can be evaluated as the corresponding definite integral: $f(3)-f(1)=\int_1^3 \frac{\sqrt{x}}{(1+x)^2}\,dx.$
Use the substitution $x=t^2, \quad dx=2t\,dt, \quad \sqrt{x}=t.$ Then when $x=1$ , $t=1$ , and when $x=3$ , $t=\sqrt{3}$ .

So,

=\int_1^{\sqrt{3}} \frac{t}{(1+t^2)^2}(2t\,dt) =\int_1^{\sqrt{3}} \frac{2t^2}{(1+t^2)^2}\,dt.$$ 3. Simplify the integrand: $$\frac{2t^2}{(1+t^2)^2} =\frac{2(1+t^2)-2}{(1+t^2)^2} =\frac{2}{1+t^2}-\frac{2}{(1+t^2)^2}.$$ Hence, $$f(3)-f(1)=\int_1^{\sqrt{3}} \frac{2}{1+t^2}\,dt-\int_1^{\sqrt{3}} \frac{2}{(1+t^2)^2}\,dt.$$ 4. Now use the standard result $$\int \frac{1}{(1+t^2)^2}\,dt=\frac{1}{2}\tan^{-1}t+\frac{t}{2(1+t^2)}+C.$$ Therefore, $$\int \frac{2}{(1+t^2)^2}\,dt=\tan^{-1}t+\frac{t}{1+t^2}+C.$$ So an antiderivative of $$\frac{2t^2}{(1+t^2)^2}$$ is $$2\tan^{-1}t-\left(\tan^{-1}t+\frac{t}{1+t^2}\right) =\tan^{-1}t-\frac{t}{1+t^2}.$$ Thus, $$f(3)-f(1)=\left[\tan^{-1}t-\frac{t}{1+t^2}\right]_1^{\sqrt{3}}.$$ 5. Evaluate at the limits: - At $t=\sqrt{3}$: $$\tan^{-1}(\sqrt{3})=\frac{\pi}{3}, \qquad \frac{\sqrt{3}}{1+3}=\frac{\sqrt{3}}{4}.$$ So the value is $$\frac{\pi}{3}-\frac{\sqrt{3}}{4}.$$ - At $t=1$: $$\tan^{-1}(1)=\frac{\pi}{4}, \qquad \frac{1}{1+1}=\frac{1}{2}.$$ So the value is $$\frac{\pi}{4}-\frac{1}{2}.$$ Therefore, $$f(3)-f(1)=\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)-\left(\frac{\pi}{4}-\frac{1}{2}\right).$$ 6. Simplify: $$f(3)-f(1)=\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}.$$ So the correct option is: $$\boxed{\text{B}}$$ with value $$\boxed{\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}}.$$ 7. Comparison with stored correct answer: - Stored correct answer: $\text{B}$ - Derived answer: $\text{B}$ They match.