Let I(x)=\int \fracd x(x-11)^\frac1113(x+15)^\frac1513 . If I(37)-I(24)=\frac14(\frac1~b^\frac113-\frac1c^\frac113), b, c \in \mathcalN , then 3(~b+c) is equal to

Indefinite Integrals PYQ — JEE Main 2025 · 23 Jan · Shift 1 · Q39 with Solution

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Correct answer: A

We need to evaluate $I(37)-I(24)=\int_{24}^{37}\frac{dx}{(x-11)^{11/13}(x+15)^{15/13}}.$
Observe the exponents suggest the substitution $u=\frac{x-11}{x+15}.$ Then $x-11=u(x+15),$ so differentiating directly, $u=\frac{x-11}{x+15}\implies \frac{du}{dx}=\frac{(x+15)-(x-11)}{(x+15)^2}=\frac{26}{(x+15)^2}.$ Hence, $dx=\frac{(x+15)^2}{26}\,du.$

Also, $x-11=u(x+15).$ Therefore the integrand becomes

=\frac{1}{\big(u(x+15)\big)^{11/13}(x+15)^{15/13}}\cdot \frac{(x+15)^2}{26}\,du.$$ Simplify the powers of $(x+15)$: $$\big(u(x+15)\big)^{11/13}(x+15)^{15/13}=u^{11/13}(x+15)^{26/13}=u^{11/13}(x+15)^2.$$ So, $$\frac{dx}{(x-11)^{11/13}(x+15)^{15/13}}=\frac{1}{26}u^{-11/13}\,du.$$ Thus, $$I(37)-I(24)=\frac{1}{26}\int_{u(24)}^{u(37)}u^{-11/13}\,du.$$ 3. Now compute the new limits. For $x=24$: $$u(24)=\frac{24-11}{24+15}=\frac{13}{39}=\frac13.$$ For $x=37$: $$u(37)=\frac{37-11}{37+15}=\frac{26}{52}=\frac12.$$ Hence, $$I(37)-I(24)=\frac{1}{26}\int_{1/3}^{1/2}u^{-11/13}\,du.$$ 4. Integrate: $$\int u^{-11/13}du=\frac{u^{2/13}}{2/13}=\frac{13}{2}u^{2/13}.$$ Therefore, $$I(37)-I(24)=\frac{1}{26}\cdot \frac{13}{2}\left[u^{2/13}\right]_{1/3}^{1/2} =\frac14\left[\left(\frac12\right)^{2/13}-\left(\frac13\right)^{2/13}\right].$$ Now, $$\left(\frac12\right)^{2/13}=\frac{1}{2^{2/13}}=\frac{1}{4^{1/13}},$$ $$\left(\frac13\right)^{2/13}=\frac{1}{3^{2/13}}=\frac{1}{9^{1/13}}.$$ So, $$I(37)-I(24)=\frac14\left(\frac{1}{4^{1/13}}-\frac{1}{9^{1/13}}\right).$$ Comparing with $$\frac14\left(\frac{1}{b^{1/13}}-\frac{1}{c^{1/13}}\right),$$ we get $$b=4,\quad c=9.$$ 5. Therefore, $$3(b+c)=3(4+9)=3\cdot 13=39.$$ 6. Checking options: - A: $39$ ✅ - B: $22$ ❌ - C: $40$ ❌ - D: $26$ ❌ So the correct option is **A**.

View written solutionFree

Correct answer: A

We need to evaluate $I(37)-I(24)=\int_{24}^{37}\frac{dx}{(x-11)^{11/13}(x+15)^{15/13}}.$
Observe the exponents suggest the substitution $u=\frac{x-11}{x+15}.$ Then $x-11=u(x+15),$ so differentiating directly, $u=\frac{x-11}{x+15}\implies \frac{du}{dx}=\frac{(x+15)-(x-11)}{(x+15)^2}=\frac{26}{(x+15)^2}.$ Hence, $dx=\frac{(x+15)^2}{26}\,du.$

Also, $x-11=u(x+15).$ Therefore the integrand becomes

=\frac{1}{\big(u(x+15)\big)^{11/13}(x+15)^{15/13}}\cdot \frac{(x+15)^2}{26}\,du.$$ Simplify the powers of $(x+15)$: $$\big(u(x+15)\big)^{11/13}(x+15)^{15/13}=u^{11/13}(x+15)^{26/13}=u^{11/13}(x+15)^2.$$ So, $$\frac{dx}{(x-11)^{11/13}(x+15)^{15/13}}=\frac{1}{26}u^{-11/13}\,du.$$ Thus, $$I(37)-I(24)=\frac{1}{26}\int_{u(24)}^{u(37)}u^{-11/13}\,du.$$ 3. Now compute the new limits. For $x=24$: $$u(24)=\frac{24-11}{24+15}=\frac{13}{39}=\frac13.$$ For $x=37$: $$u(37)=\frac{37-11}{37+15}=\frac{26}{52}=\frac12.$$ Hence, $$I(37)-I(24)=\frac{1}{26}\int_{1/3}^{1/2}u^{-11/13}\,du.$$ 4. Integrate: $$\int u^{-11/13}du=\frac{u^{2/13}}{2/13}=\frac{13}{2}u^{2/13}.$$ Therefore, $$I(37)-I(24)=\frac{1}{26}\cdot \frac{13}{2}\left[u^{2/13}\right]_{1/3}^{1/2} =\frac14\left[\left(\frac12\right)^{2/13}-\left(\frac13\right)^{2/13}\right].$$ Now, $$\left(\frac12\right)^{2/13}=\frac{1}{2^{2/13}}=\frac{1}{4^{1/13}},$$ $$\left(\frac13\right)^{2/13}=\frac{1}{3^{2/13}}=\frac{1}{9^{1/13}}.$$ So, $$I(37)-I(24)=\frac14\left(\frac{1}{4^{1/13}}-\frac{1}{9^{1/13}}\right).$$ Comparing with $$\frac14\left(\frac{1}{b^{1/13}}-\frac{1}{c^{1/13}}\right),$$ we get $$b=4,\quad c=9.$$ 5. Therefore, $$3(b+c)=3(4+9)=3\cdot 13=39.$$ 6. Checking options: - A: $39$ ✅ - B: $22$ ❌ - C: $40$ ❌ - D: $26$ ❌ So the correct option is **A**.