Choose the correct- answer-10 x 9-10xlog e 10- x 10-10x dx equals a 10- x - x -10- Cllb 10- x - x -10- Cllc 10- x - x -10-1- Cd -log-10- x-x-10-C l

∫ ( 10 x^9+10^x log_e 10/x^10+10^x )dx Let x^10+10^x =t ⇒ (10x^9+10^x log_e 10)dx=dt ⇒ dx =dt/10 x^9+10^x log_e 10 ∴∫ ( 10x^9+10^x log_e 10)/x^10+10^x )dx =∫10x ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

Choose the co...

Question

Choose the correct. answer.
$\int (\frac{10 x^{9} + 10^{x} l o g_{e} 10}{x^{10} + 10^{x}}) d x$ equals
$(a) 10^{x} + x^{10} + C (b) 10^{x} + x^{10} + C (c) (10^{x} - x^{10})^{- 1} + C (d) l o g | 10^{x} + x^{10} | + C$

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Solution

$\int (\frac{10 x^{9} + 10^{x} l o g_{e} 10}{x^{10} + 10^{x}}) d x$
Let $x^{10} + 10^{x} = t \Rightarrow (10 x^{9} + 10^{x} l o g_{e} 10) d x = d t$
$\Rightarrow d x = \frac{d t}{10 x^{9} + 10^{x} l o g_{e} 10} ∴ \int (\frac{10 x^{9} + 10^{x} l o g_{e} 10)}{x^{10} + 10^{x}}) d x = \int \frac{10 x^{9} + 10^{x} l o g_{e} 10}{t} \times \frac{d t}{10 x^{9} + 10^{x} l o g_{e} 10} = \int \frac{d t}{t} = l o g | t | + C = l o g | 10^{x} + x^{10} | + C$
So, (d) is the correct option.

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Similar questions

\int \frac{10 x^{9} + 10^{x} {log}_{e^{10}} d x}{x^{10} + 10^{x}}

equals

Q. Given that

{log}_{10} sin x + {log}_{10} cos x = - 1

and that

{log}_{10} (sin x + cos x) = \frac{1}{2} ({log}_{10} n - 1) .

Find

n .

(correct answer + 3, wrong answer 0)

Q. If

y = {log}_{10} x + {log}_{x} 10 + {log}_{x} x + {log}_{10} 10

, then

\frac{d y}{d x}

is equal to

$Find x if$ $l o g_{10} (x + 1) + l o g_{10} (x - 1) = l o g_{10} 11 + 2 l o g_{10} 3$

Q. The inverse of the function is

f (x) = \frac{10^{x} - 10^{- x}}{10^{x} + 10^{- x}} + 1

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Choose the correct. answer. ∫(10x9+10xloge10x10+10x)dx equals (a)10x+x10+C(b)10x+x10+C(c)(10x−x10)−1+C(d)log|10x+x10|+C

∫(10x9+10xloge10x10+10x)dx Let x10+10x=t⇒(10x9+10xloge10)dx=dt ⇒dx=dt10x9+10xloge10∴∫(10x9+10xloge10)x10+10x)dx=∫10x9+10xloge10t×dt10x9+10xloge10=∫dtt=log|t|+C=log|10x+x10|+C So, (d) is the correct option.

Choose the correct. answer.
$\int (\frac{10 x^{9} + 10^{x} l o g_{e} 10}{x^{10} + 10^{x}}) d x$ equals
$(a) 10^{x} + x^{10} + C (b) 10^{x} + x^{10} + C (c) (10^{x} - x^{10})^{- 1} + C (d) l o g | 10^{x} + x^{10} | + C$