Q103 / 180 · COMEDK 2025 (10 May Shift 1) · Mathematics
$\int \log x^2 d x=$
- A$\log x^2+x+c$
- Correct answerB$x \log x^2-2 x+c$
- C$x \log x^2+x+c$
- D$x \log x^2-1+c$
Solution
The integral is $I = \int \log(x^2) dx$.
Using the property of logarithms, $\log(x^2) = 2 \log x$, so $I = 2 \int \log x dx$.
Using integration by parts, $\int u dv = uv - \int v du$, where $u = \log x$ and $dv = dx$.
Then $du = \dfrac{1}{x} dx$ and $v = x$.
Therefore, $I = 2 [x \log x - \int x \cdot \dfrac{1}{x} dx] = 2 [x \log x - \int 1 dx] = 2 [x \log x - x] + c = 2x \log x - 2x + c$.
Since $2x \log x = x(2 \log x) = x \log(x^2)$, we have $I = x \log(x^2) - 2x + c$.
Answer: $x \log x^2-2 x+c$
Using the property of logarithms, $\log(x^2) = 2 \log x$, so $I = 2 \int \log x dx$.
Using integration by parts, $\int u dv = uv - \int v du$, where $u = \log x$ and $dv = dx$.
Then $du = \dfrac{1}{x} dx$ and $v = x$.
Therefore, $I = 2 [x \log x - \int x \cdot \dfrac{1}{x} dx] = 2 [x \log x - \int 1 dx] = 2 [x \log x - x] + c = 2x \log x - 2x + c$.
Since $2x \log x = x(2 \log x) = x \log(x^2)$, we have $I = x \log(x^2) - 2x + c$.
Answer: $x \log x^2-2 x+c$
Hence, the correct answer is option: 'B'.