Explanation for the correct option:Finding the value of ∫{log√(x)/3x}dx:Consider the given Equation as,I=∫{log√(x)/3x}dx0ex0exI=1/3∫{log√(x)/x}dxLet us assume t ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫log√x/3xdx=

Question

$\int \{\frac{\log \sqrt{x}}{3 x}\} d x =$

$(\frac{1}{3}) {(\log \sqrt{x})}^{2} + c$

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$(\frac{2}{3}) {(\log \sqrt{x})}^{2} + c$

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$(\frac{2}{3}) {(\log x)}^{2} + c$

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$(\frac{1}{3}) {(\log x)}^{2} + c$

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Solution

The correct option is A
$(\frac{1}{3}) {(\log \sqrt{x})}^{2} + c$

Explanation for the correct option:
Finding the value of $\int \{\frac{\log \sqrt{x}}{3 x}\} d x$ :
Consider the given Equation as,
$I = \int \{\frac{\log \sqrt{x}}{3 x}\} d x I = \frac{1}{3} \int \{\frac{\log \sqrt{x}}{x}\} d x$
Let us assume that
$\log \sqrt{x} = t \Rightarrow \frac{1}{\sqrt{x}} \times \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{1}{x} d x = 2 d t$
Then $I$ becomes,
$I = \frac{1}{3} \int \{\frac{\log \sqrt{x}}{x}\} d x = \frac{2}{3} \int t d t = \frac{2}{3} \frac{t^{2}}{2} + c I = \frac{1}{3} t^{2} + c$
Substituting the value of $t$ the above equation becomes,
$I = \frac{1}{3} {(\log \sqrt{x})}^{2} + c$
Hence, the correct answer is Option (A).

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∫logx3xdx=

$\int \{\frac{\log \sqrt{x}}{3 x}\} d x =$