Question

$\int {\left(\log x\right)}^{2}dx=x{\left[f\left(x\right)\right]}^2+Ax\left[f\left(x\right)-1\right]+c$, then

• A. $f\left(x\right)=\log x,A=2$
• B. $f\left(x\right)=\log x,A=-2$
• C. $f\left(x\right)=-\log x,A=2$
• D. $f\left(x\right)=-\log x,A=-2$

Answer 1

The correct option is B

$f\left(x\right)=\log x,A=-2$

Explanation for the correct option:

Finding the value using the substitution method:

Consider the given equation,

$\int {\left(\log x\right)}^{2}dx=x{\left[f\left(x\right)\right]}^2+Ax\left[f\left(x\right)-1\right]+c......\left(1\right)$

$\int {\left(\log x\right)}^{2}dx=\int {\left(\log x\right)}^2.1.dx$

We know that,

$\int uvdx=u\int v-\int \left(du\int vdx\right)dx$

$$\begin{gathered} \int {\left(\log x\right)}^2.1.dx={\left(\log x\right)}^{2}x-\int 2\log x\times \frac{1}{x}\times xdx \\ \int {\left(\log x\right)}^2.1.dx=x{\left(\log x\right)}^2-2\left[x\log x-\int \frac{1}{x}\times xdx\right] \\ \int {\left(\log x\right)}^2.1.dx=x{\left(\log x\right)}^2-2\left[x\log x-x\right]+c \\ \int {\left(\log x\right)}^2.1.dx=x{\left(\log x\right)}^2-2x\left[\log x-1\right]+c.......\left(2\right) \end{gathered}$$

Comparing equations $\left(1\right)$ and $\left(2\right)$ we get

$x{\left(\log x\right)}^2-2x\left[\log x-1\right]+c=x{\left[f\left(x\right)\right]}^2+Ax\left[f\left(x\right)-1\right]+c$

Then,

$$\begin{gathered} f\left(x\right)=\log x \\ A=-2 \end{gathered}$$

Hence, the correct answer is Option (B).