Question

Evaluate the integral: ${\int }_1^2\log xdx$

• (A) $2\log 2-1$
• (B) $2\log 2+1$
• (C) $2\log 2-3$
• (D) None of these

Answer 1

The Correct Option is A

Explanation:
Let's first integrate the expression under integral by parts.$I=\int \log xdx=\int (1)(\log x)dx$Considering $\log x$ as the first function and 1 as the second function, we get:$=(\log x)\int 1dx-\int [\frac{1}{x}\int 1dx]dx$ [$\because \int f(x)g(x)dx=f(x)\int g(x)dx-\int [f^{\prime }(x)\int g(x)dx]dx$$=(\log x)x-x+C$Putting the limits of the definite integral, we get:${\int }_1^2\log x\mathrm{d}x$$=[x(\log x)-x{]}_1^2$$=(2\log 2-2)-(0-1)$$=2\log 2-1$Hence, the correct option is (A).