Question

In some of the cases we can split the integrand into the sum of the two functions such that the integration of one of them by parts produces an integral which cancels the other integral. Suppose we have an integral of the type
$\int \left[f\right(x\left)h\right(x)+g(x\left)\right]dx$
Let $\int f\left(x\right)h\left(x\right)dx=I_1$ and $\int g\left(x\right)dx=I_2$
Integrating $I_1$ by parts, we get
$I_1=f\left(x\right)\int h\left(x\right)dx-\int \left\{f^{\prime }\right(x)\int h(x\left)dx\right\}dx$
Now find $\int (\frac{1}{logx}-\frac{1}{(logx{)}^2})dx$ (x > 0)

• A. $log\left(logx\right)+c$
  
  
  
  
• B. $xlogx+c$
• C. $\frac{x}{logx}+c$
• D. $x+logx+c$

Answer 1

The correct option is C

$\frac{x}{logx}+c$

$I=\int (\frac{1}{lo{g}_{e}x}-\frac{1}{(lo{g}_{e}x{)}^2})dx$
Put $lo{g}_{e}x=t$
$\Rightarrow x=e^t$
$\therefore dx=e^{t}dt$
$I=\int (\frac{1}{t}-\frac{1}{t^2})e^{t}dt=e^t.\frac{1}{t}+C=\frac{x}{logx}+C$