Question

Evaluate $\int \left[\right(\log x{)}^2-7\log x$ $+9]dx$

• A. $x\left[\right(\log x{)}^2-9\log x]+c$
• B. $x\left[\right(\log x{)}^2-9\left(\log x\right)+18]+c$
• C. $x\left[\right(\log x{)}^2+9\log x-18]+c$
• D. $x\left[\right(\log x{)}^2-7\log x+9]+c$

Answer 1

The correct option is B $x\left[\right(\log x{)}^2-9\left(\log x\right)+18]+c$

$\int \left[\right(\log x{)}^2-7\left(\log x\right)+9]dx$

We know, $\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

Now by integrating the functions by parts, we get

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

$\int \left(\log x\right)dx=x\log x-x$

$\therefore \int \left[\right[\left(\log x{)}^2\right]-7\left(\log x\right)+9]dx$

$=x(\log x{)}^2-2\int (\log x)-7\int (\log x)dx+\int 9dx+c$

$=x(\log x{)}^2-9[x\log x-x]+9x+c$

$=x(\log x{)}^2-9x\log x+18x+c$

$x\left[\right(\log x{)}^2-9\left(\log x\right)+18]+c$