JEE Main 2026 — Mathematics Calculus

The given integral is $\displaystyle f(x) = \int \left(\dfrac{16x + 24}{x^2 + 2x - 15}\right) dx$f(x)=∫(x2+2x−1516x+24​)dx.

Factorizing the denominator, we get $x^2 + 2x - 15 = (x + 5)(x - 3)$x2+2x−15=(x+5)(x−3).

Using partial fractions, we can write:

$$\dfrac{16x + 24}{(x + 5)(x - 3)} = \dfrac{A}{x + 5} + \dfrac{B}{x - 3}$$(x+5)(x−3)16x+24​=x+5A​+x−3B​

$$16x + 24 = A(x - 3) + B(x + 5)$$16x+24=A(x−3)+B(x+5)

Substituting $x = 3$x=3, we get $72 = 8B \Rightarrow B = 9$72=8B⇒B=9.

Substituting $x = -5$x=−5, we get $-56 = -8A \Rightarrow A = 7$−56=−8A⇒A=7.

Therefore, the integral becomes:

$$f(x) = \int \left(\dfrac{7}{x + 5} + \dfrac{9}{x - 3}\right) dx$$f(x)=∫(x+57​+x−39​)dx

$$f(x) = 7 \log_e|x + 5| + 9 \log_e|x - 3| + C$$f(x)=7loge​∣x+5∣+9loge​∣x−3∣+C

Given $f(4) = 14 \log_e(3)$f(4)=14loge​(3), we substitute $x = 4$x=4:

$$f(4) = 7 \log_e(9) + 9 \log_e(1) + C = 14 \log_e(3) + C$$f(4)=7loge​(9)+9loge​(1)+C=14loge​(3)+C

Since $f(4) = 14 \log_e(3)$f(4)=14loge​(3), we get $C = 0$C=0.

Thus, $f(x) = 7 \log_e|x + 5| + 9 \log_e|x - 3|$f(x)=7loge​∣x+5∣+9loge​∣x−3∣.

Now, substituting $x = 7$x=7:

$$f(7) = 7 \log_e(12) + 9 \log_e(4)$$f(7)=7loge​(12)+9loge​(4)

$$f(7) = 7 \log_e(2^2 \cdot 3) + 9 \log_e(2^2)$$f(7)=7loge​(22⋅3)+9loge​(22)

$$f(7) = 7(2 \log_e(2) + \log_e(3)) + 18 \log_e(2)$$f(7)=7(2loge​(2)+loge​(3))+18loge​(2)

$$f(7) = 14 \log_e(2) + 7 \log_e(3) + 18 \log_e(2)$$f(7)=14loge​(2)+7loge​(3)+18loge​(2)

$$f(7) = 32 \log_e(2) + 7 \log_e(3)$$f(7)=32loge​(2)+7loge​(3)

$$f(7) = \log_e(2^{32} \cdot 3^7)$$f(7)=loge​(232⋅37)

Comparing this with $f(7) = \log_e(2^{\alpha} \cdot 3^{\beta})$f(7)=loge​(2α⋅3β), we get $\alpha = 32$α=32 and $\beta = 7$β=7.

Therefore, $\alpha + \beta = 32 + 7 = 39$α+β=32+7=39.

Answer: $39$39