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Choose the correct answer in $\Large \int \normalsize \frac{10x^9+10^xlog_{e^{10}}dx}{x^{10}+{10}^x}$ equals

$\begin{array}{1 1} (A)\;{10}^x-x^{10}+C \\ (B)\;{10}^x+x^{10}+C \\ (C)\;({10}^x-x^{10})^{-1}+C \\ (D)\;log({10}^x+x^{10})+C\end{array}$

1 Answer

  • (i)Method of substitution :
  • Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
  • $\Rightarrow $dx=g'(t)dt.
  • Thus $I=\int f(g(t).g'(t))dt.$
  • (ii)$x^ndx=\frac{x^{n+1}}{n+1}+c.$
Given $I=\int \frac{10x^9+10x^xlog_e10}{x^{10}+10^x}dx$.
Put $x^{10}+10^x=t.$
On differentiating this,
Now substituting of x and dx we get,
$I=\int \frac{dt}{t}$.
On integrating we get,
$log t+c$.
Substituting back for t we get,
$\int \frac{10x^9+10x^xlog_e10}{x^{10}+10^x}dx=log (x^{10}+10^x)+c$.
Hence D is the correct answer.


answered Jan 30, 2013 by sreemathi.v