_ ^ 10 x^9 + 10 ^x _e 10 10 ^x + x^ 10 ;dx = - Vidyadip

MCQ

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

A
$ - \frac{1}{2}\frac{1}{{{{({{10}^x} + {x^{10}})}^2}}} + c$
✓
$\log ({10^x} + {x^{10}}) + c$
C
$\frac{1}{2}\frac{1}{{{{({{10}^x} + {x^{10}})}^2}}} + c$
D
None of these

✓

Answer

Correct option: B.

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

b
(b) Put ${x^{10}} + {10^x} = t \Rightarrow (10{x^9} + {10^x}{\log _e}10)\,dx = dt,$
then $\int_{}^{} {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}\,dx} = \int_{}^{} {\frac{1}{t}\,dt = \log t + c} $
$ = \log ({x^{10}} + {10^x}) + c.$

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