Q82 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics
$\displaystyle\int \dfrac{30x^{14} + 15^x \log 225}{x^{15} + 15^x} \, dx =$
- A$x^{15} + 15x^{14} + 30x^{13} + c$, where c is the constant of integration
- B$\dfrac{\log(x^{15} + 15^x)}{2} + c$, where c is the constant of integration
- Correct answerC$\log(x^{15} + 15^x)^2 + c$, where c is the constant of integration
- D$\log(x^{15} + 15^x) + c$, where c is the constant of integration
Solution
Let $I = \displaystyle\int \dfrac{30x^{14} + 15^x \log 225}{x^{15} + 15^x} \, dx$
Let $x^{15} + 15^x = t$
Differentiating both sides with respect to $x$, we get
$(15x^{14} + 15^x \log 15) \, dx = dt$
Multiplying by $2$, we get
$(30x^{14} + 2 \cdot 15^x \log 15) \, dx = 2 \, dt$
Since $2 \log 15 = \log(15^2) = \log 225$, this becomes
$(30x^{14} + 15^x \log 225) \, dx = 2 \, dt$
Substituting this into the integral, we obtain
$I = \displaystyle\int \dfrac{2}{t} \, dt$
$I = 2 \log t + c$
$I = \log(t^2) + c$
$I = \log(x^{15} + 15^x)^2 + c$
Answer: $\log(x^{15} + 15^x)^2 + c$, where c is the constant of integration
Let $x^{15} + 15^x = t$
Differentiating both sides with respect to $x$, we get
$(15x^{14} + 15^x \log 15) \, dx = dt$
Multiplying by $2$, we get
$(30x^{14} + 2 \cdot 15^x \log 15) \, dx = 2 \, dt$
Since $2 \log 15 = \log(15^2) = \log 225$, this becomes
$(30x^{14} + 15^x \log 225) \, dx = 2 \, dt$
Substituting this into the integral, we obtain
$I = \displaystyle\int \dfrac{2}{t} \, dt$
$I = 2 \log t + c$
$I = \log(t^2) + c$
$I = \log(x^{15} + 15^x)^2 + c$
Answer: $\log(x^{15} + 15^x)^2 + c$, where c is the constant of integration
Hence, the correct answer is option: 'C'.