Question

$\int \frac{10x^9 + 10^x log_e 10\ dx}{x^{10}+10^x}$ equals,
(A) $10^x - x^{10} + C$
(B) $10^x + x^{10} + C$
(C) $(10^x - x^{10})-1 + C$
(D) $log (10^x + x^{10}) + C$

Answer 1

Answer:

Option D is correct.

Step-by-step explanation:

To integrate the given function we can apply substitution Method.

$\int \frac{10x^9 + 10^x log_e 10\ dx}{x^{10}+10^x} \\ \\ let \\ \\ x^{10}+10^x = t \\ \\ differentiate \: both \: side \\ \\ (10 {x}^{9} + {10}^{x} log_e 10)dx = dt \\ \\ \int \frac{1}{t}dt = log \: t + C\\ \\ undo \: substitution \\ \\ \int \frac{10x^9 + 10^x log_e 10\ dx}{x^{10}+10^x} = log( \: {x^{10}+10^x} )+ C$

Option D is correct.

Hope it helps you.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{ \int \frac{ dx}{x} = \log x\: + c}$

TO DETERMINE

$\displaystyle \sf{ \int \frac{(10 {x}^{9} + {10}^{x} log_{e}10) dx}{ {x}^{10} + {10}^{x} }}$

CALCULATION

$\displaystyle \sf{ \int \frac{(10 {x}^{9} + {10}^{x} log_{e}10) dx}{ {x}^{10} + {10}^{x} }}$

Let

$\displaystyle \sf{ z = {x}^{10} + {10}^{x} }$

$\implies \displaystyle \sf{ \frac{dz}{dx} = {10 {x}^{9} + {10}^{x} log_{e}10 }}$

$\implies \displaystyle \sf{ {dz} = ( {10 {x}^{9} + {10}^{x} log_{e}10) }}dx$

Hence

$\displaystyle \sf{ \int \frac{(10 {x}^{9} + {10}^{x} log_{e}10) dx}{ {x}^{10} + {10}^{x} }}$

$= \displaystyle \sf{ \int \frac{dz}{z} }$

$= \sf{ \log z \: + c \: \: ( \:where \: c \: is \: constant) \: }$

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

RESULT

$\boxed{ \displaystyle \sf{ \int \frac{(10 {x}^{9} + {10}^{x} log_{e}10) dx}{ {x}^{10} + {10}^{x} }} \: = \: { \log ( \: {x}^{10} + {10}^{x} \: ) \: + c \: }}$

━━━━━━━━━━━━━━━━

LEARN MORE FROM BRAINLY

Use Rolles Theorem to find a real number in the interval (0,3) Where the derivative of the function f(x)=x(x-3)^2 Vanish

https://brainly.in/question/22736169