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Q38  Choose the correct answer \int \frac{10 x^ 9 + 10 ^x \log _ e 10 dx }{x ^{10}+ 10 ^x }dx\: \: \: equals

(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

 

Answers (1)

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Given integral \int \frac{10 x^ 9 + 10 ^x \log _ e 10 dx }{x ^{10}+ 10 ^x }dx

Taking the denominator x^{10} +10^x = t

Now differentiating both sides we get

\therefore \left ( 10x^9+10^x\log_{e}10 \right )dx = dt

\implies \int \frac{10x^9+10^x\log_{e}10}{x^{10}+10^x} dx = \int \frac{dt}{t}

= \log t +C

Back substituting the value of t,

= \log (x^{10}+10^x) +C

Therefore the correct answer is D.

Posted by

Divya Prakash Singh

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