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

log | 10^{x} + x^{10} | + c

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- log | 10^{x} + x^{10} | + c

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10^{x} + x^{10} + c

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log | 10^{x} + x^{9} | + c

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Solution

The correct option is C $log | 10^{x} + x^{10} | + c$
Consider $N (x) = 10 x^{9} + 10^{x} {log}_{e} 10$ and $D (x) = 10^{x} + x^{10}$
$\Rightarrow D^{'} (x) = 10^{x} {log}_{e} 10 + 10 x^{9}$
$\Rightarrow N (x) = D^{'} (x)$
Thus, we have $\int \frac{N (x)}{D (x)} d x = \int \frac{d (D (x))}{D (x)}$
$= log [D (x)] + c$

$∴ \int \frac{10 x^{9} + 10^{x} {log}_{e} 10}{10^{x} + x^{10}} d x = log ∣ ∣ 10^{x} + x^{10} ∣ ∣ + c$

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Similar questions

Q. Assertion :

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

Reason:

\int \frac{f^{'} (x)}{f (x)} d x = log | f (x) | + C

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

equals

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

equals

Q. Find

\frac{d y}{d x}

for

y = x^{10} + 10^{x} + 10 x + 10

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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∫10x9+10xloge10x10+10xdx=

The correct option is C log|10x+x10|+cConsider N(x)=10x9+10xloge10 and D(x)=10x+x10⇒D′(x)=10xloge10+10x9⇒N(x)=D′(x)Thus, we have ∫N(x)D(x)dx=∫d(D(x))D(x)=log[D(x)]+c∴∫10x9+10xloge1010x+x10dx=log∣∣10x+x10∣∣+c

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