Indefinite IntegrationmediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Differential Equation Solution: x(e^2) = 2e^2/3 | JEE Main 2026

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Question
Let f(x)=∫(16x+24x2+2x−15)dxf(x)=\displaystyle\int\left(\dfrac{16x+24}{x^2+2x-15}\right)dx. If f(4)=14log⁡e3f(4)=14\log_e 3 and f(7)=log⁡e ⁣(2α⋅3β)f(7)=\log_e\!\left(2^{\alpha}\cdot 3^{\beta}\right), α,β∈N\alpha,\beta\in\mathbb{N}, then α+β\alpha+\beta is equal to
A3131
B3737
C3939correct
D4141
Solution
Step 1: Split the numerator so one part is the derivative of the denominator: 16x+24=8(2x+2)+816x+24=8(2x+2)+8, and x2+2x−15=(x+5)(x−3)x^2+2x-15=(x+5)(x-3). So
f(x)=∫8(2x+2)x2+2x−15 dx+∫8(x+5)(x−3) dx.f(x)=\int\frac{8(2x+2)}{x^2+2x-15}\,dx+\int\frac{8}{(x+5)(x-3)}\,dx.
 Step 2: The first integral is 8ln⁡∣x2+2x−15∣8\ln|x^2+2x-15|. For the second, 8(x+5)(x−3)=1x−3−1x+5\dfrac{8}{(x+5)(x-3)}=\dfrac{1}{x-3}-\dfrac{1}{x+5}, giving ln⁡∣x−3x+5∣\ln\left|\dfrac{x-3}{x+5}\right|. Hence
f(x)=8ln⁡∣x2+2x−15∣+ln⁡∣x−3x+5∣+C.f(x)=8\ln|x^2+2x-15|+\ln\left|\frac{x-3}{x+5}\right|+C.
 Step 3: Evaluate at x=4x=4: x2+2x−15=9x^2+2x-15=9 and x−3x+5=19\dfrac{x-3}{x+5}=\dfrac19, so
f(4)=8ln⁡9+ln⁡19+C=8ln⁡9−ln⁡9+C=7ln⁡9+C=14ln⁡3+C.f(4)=8\ln9+\ln\tfrac19+C=8\ln9-\ln9+C=7\ln9+C=14\ln3+C.
 Given f(4)=14ln⁡3⇒C=0f(4)=14\ln3\Rightarrow C=0. Step 4: Evaluate at x=7x=7: x2+2x−15=48x^2+2x-15=48 and x−3x+5=412=13\dfrac{x-3}{x+5}=\dfrac{4}{12}=\dfrac13, so
f(7)=8ln⁡48+ln⁡13=ln⁡4883.f(7)=8\ln48+\ln\tfrac13=\ln\frac{48^8}{3}.
 Step 5: Since 48=24⋅348=2^4\cdot3, 488=232⋅3848^8=2^{32}\cdot3^{8}, so
f(7)=ln⁡232⋅383=ln⁡ ⁣(232⋅37).f(7)=\ln\frac{2^{32}\cdot3^{8}}{3}=\ln\!\left(2^{32}\cdot3^{7}\right).
 Thus α=32, β=7\alpha=32,\ \beta=7 and α+β=39\alpha+\beta=39. Correct answer: (3)
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