Browse Questions

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

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

Toolbox:
• (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,

$(10x^9+10^xlog_e10)dx=dt.$

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.