The value of the integral \int\limits_0^1 \sqrt x dx \over (1 + x)(1 + 3x)(3 + x) is :

\pi \over 8( 1 - \sqrt 3 \over 2 )

\pi \over 4( 1 - \sqrt 3 \over 6 )

\pi \over 8( 1 - \sqrt 3 \over 6 )

\pi \over 4( 1 - \sqrt 3 \over 2 )

Definite Integration PYQ — JEE Main 2021 · 27 Aug · Shift 2 · Q34 with Solution

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Correct answer: NONE OF THE GIVEN OPTIONS; THE CORRECT VALUE IS $\FRAC{\PI}{8}(\SQRT{3}-1)$

We need to evaluate $I=\int_0^1 \frac{\sqrt{x}\,dx}{(1+x)(1+3x)(3+x)}.$
Use the substitution $x=t^2,\qquad dx=2t\,dt,\qquad \sqrt{x}=t.$ As $x$ goes from $0$ to $1$ , $t$ goes from $0$ to $1$ .

=\int_0^1 \frac{2t^2}{(1+t^2)(1+3t^2)(3+t^2)}\,dt.$$ 3. Now decompose into partial fractions of the form $$\frac{2t^2}{(1+t^2)(1+3t^2)(3+t^2)} =\frac{A}{1+t^2}+\frac{B}{1+3t^2}+\frac{C}{3+t^2}.$$ Multiplying through by $(1+t^2)(1+3t^2)(3+t^2)$, $$2t^2=A(1+3t^2)(3+t^2)+B(1+t^2)(3+t^2)+C(1+t^2)(1+3t^2).$$ Expand: $$A(1+3t^2)(3+t^2)=A(3+10t^2+3t^4),$$ $$B(1+t^2)(3+t^2)=B(3+4t^2+t^4),$$ $$C(1+t^2)(1+3t^2)=C(1+4t^2+3t^4).$$ Hence $$2t^2=(3A+3B+C)+(10A+4B+4C)t^2+(3A+B+3C)t^4.$$ Comparing coefficients: $$3A+3B+C=0,$$ $$10A+4B+4C=2,$$ $$3A+B+3C=0.$$ Solving, $$A=-\frac12,\qquad B=\frac34,\qquad C=\frac34.$$ Thus $$I=\int_0^1\left(-\frac{1}{2(1+t^2)}+\frac{3}{4(1+3t^2)}+\frac{3}{4(3+t^2)}\right)dt.$$ 4. Integrate term by term. We use: $$\int \frac{dt}{1+t^2}=\tan^{-1}t,$$ $$\int \frac{dt}{1+3t^2}=\frac{1}{\sqrt3}\tan^{-1}(\sqrt3 t),$$ $$\int \frac{dt}{3+t^2}=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{t}{\sqrt3}\right).$$ Therefore, $$I=\left[-\frac12\tan^{-1}t+\frac{3}{4\sqrt3}\tan^{-1}(\sqrt3 t)+\frac{3}{4\sqrt3}\tan^{-1}\left(\frac{t}{\sqrt3}\right)\right]_0^1.$$ At $t=1$: $$\tan^{-1}(1)=\frac\pi4, \qquad \tan^{-1}(\sqrt3)=\frac\pi3, \qquad \tan^{-1}\left(\frac1{\sqrt3}\right)=\frac\pi6.$$ At $t=0$, all terms are $0$. So $$I=-\frac12\cdot\frac\pi4+\frac{3}{4\sqrt3}\left(\frac\pi3+\frac\pi6\right).$$ Now $$\frac\pi3+\frac\pi6=\frac\pi2,$$ so $$I=-\frac\pi8+\frac{3}{4\sqrt3}\cdot\frac\pi2 =-\frac\pi8+\frac{3\pi}{8\sqrt3}.

Since $\frac{3}{\sqrt3}=\sqrt3,$ we get $I=\frac\pi8(\sqrt3-1).$

Our value $\frac\pi8(\sqrt3-1)$ does not match any of the given options.

So the stored answer A is not correct.

View written solutionFree

Correct answer: NONE OF THE GIVEN OPTIONS; THE CORRECT VALUE IS $\FRAC{\PI}{8}(\SQRT{3}-1)$

We need to evaluate $I=\int_0^1 \frac{\sqrt{x}\,dx}{(1+x)(1+3x)(3+x)}.$
Use the substitution $x=t^2,\qquad dx=2t\,dt,\qquad \sqrt{x}=t.$ As $x$ goes from $0$ to $1$ , $t$ goes from $0$ to $1$ .

=\int_0^1 \frac{2t^2}{(1+t^2)(1+3t^2)(3+t^2)}\,dt.$$ 3. Now decompose into partial fractions of the form $$\frac{2t^2}{(1+t^2)(1+3t^2)(3+t^2)} =\frac{A}{1+t^2}+\frac{B}{1+3t^2}+\frac{C}{3+t^2}.$$ Multiplying through by $(1+t^2)(1+3t^2)(3+t^2)$, $$2t^2=A(1+3t^2)(3+t^2)+B(1+t^2)(3+t^2)+C(1+t^2)(1+3t^2).$$ Expand: $$A(1+3t^2)(3+t^2)=A(3+10t^2+3t^4),$$ $$B(1+t^2)(3+t^2)=B(3+4t^2+t^4),$$ $$C(1+t^2)(1+3t^2)=C(1+4t^2+3t^4).$$ Hence $$2t^2=(3A+3B+C)+(10A+4B+4C)t^2+(3A+B+3C)t^4.$$ Comparing coefficients: $$3A+3B+C=0,$$ $$10A+4B+4C=2,$$ $$3A+B+3C=0.$$ Solving, $$A=-\frac12,\qquad B=\frac34,\qquad C=\frac34.$$ Thus $$I=\int_0^1\left(-\frac{1}{2(1+t^2)}+\frac{3}{4(1+3t^2)}+\frac{3}{4(3+t^2)}\right)dt.$$ 4. Integrate term by term. We use: $$\int \frac{dt}{1+t^2}=\tan^{-1}t,$$ $$\int \frac{dt}{1+3t^2}=\frac{1}{\sqrt3}\tan^{-1}(\sqrt3 t),$$ $$\int \frac{dt}{3+t^2}=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{t}{\sqrt3}\right).$$ Therefore, $$I=\left[-\frac12\tan^{-1}t+\frac{3}{4\sqrt3}\tan^{-1}(\sqrt3 t)+\frac{3}{4\sqrt3}\tan^{-1}\left(\frac{t}{\sqrt3}\right)\right]_0^1.$$ At $t=1$: $$\tan^{-1}(1)=\frac\pi4, \qquad \tan^{-1}(\sqrt3)=\frac\pi3, \qquad \tan^{-1}\left(\frac1{\sqrt3}\right)=\frac\pi6.$$ At $t=0$, all terms are $0$. So $$I=-\frac12\cdot\frac\pi4+\frac{3}{4\sqrt3}\left(\frac\pi3+\frac\pi6\right).$$ Now $$\frac\pi3+\frac\pi6=\frac\pi2,$$ so $$I=-\frac\pi8+\frac{3}{4\sqrt3}\cdot\frac\pi2 =-\frac\pi8+\frac{3\pi}{8\sqrt3}.

Since $\frac{3}{\sqrt3}=\sqrt3,$ we get $I=\frac\pi8(\sqrt3-1).$

Our value $\frac\pi8(\sqrt3-1)$ does not match any of the given options.

So the stored answer A is not correct.