int {frac{{10{x^9} + {{10}^x}{{log }_e}10}}{{ | vedclass

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(B) Let $I = \int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}} \;dx$. Substitute $t = {x^{10}} + {10^x}$. Differentiating with re

Vedclass · Accepted Answer

$\log ({10^x} + {x^{10}}) + c$

Vedclass · Answer

(B) Let $I = \int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}} \;dx$. Substitute $t = {x^{10}} + {10^x}$. Differentiating with respect to $x$,we get $dt = (10{x^9} + {10^x}{\log _e}10) \;dx$. Substituting these into the integral,we get $I = \int {\frac{1}{t}} \;dt$. Integrating,we get $I = \log |t| + c$. Substituting back the value of $t$,we get $I = \log |{x^{10}} + {10^x}| + c$.

$\int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}} \;dx = $

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