Question
$\int \frac{10 x^{9}+10^{x} \log _{e}^{10} d x}{x^{10}+10^{x}}$ बराबर है:

Answer

$\int\left(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\right) d x$
माना $x^{10} + 10^x = t$
$\Rightarrow (10x^9 + 10^x\log_e 10) dx = dt$
$\Rightarrow dx = \frac{d t}{10 x^{9}+10^{x} \log _{e} 10}$
$\therefore \int\left(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\right) d x$
$=\int \frac{10 x^{9}+10^{x} \log _{e} \cdot 10}{t} \times \frac{d t}{10 x^{9}+10^{x} \log _{e} 10}$
$=\int \frac{d t}{t}= \log |t| + C $
$= \log ∣10^x + x^{10}∣ + C$

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