-1-x-3-1-x2-3 x d x-C- x-3

The correct option is D 23log(√(x)(x 3))∫[ 1x 3 1x(x 3) ]dx =∫[ 1x 3+13x 13(x 3) ] dx =∫[ 13x+23(x 3) ]dx nbsp; = 13log x +23log(x 3)+C =23log√(x)+ 23log ...

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Byju's Answer

Standard XII

Mathematics

Volume of Parallelopiped

∫1/x-3-1/x2-3...

Question

$\int (\begin{matrix} \frac{1}{x - 3} - \frac{1}{x^{2} - 3 x} \end{matrix}) d x =$ ____ $+ C, x > 3$

\frac{1}{3} log (x (x - 3))

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\frac{1}{3} log (\sqrt{x} (x - 3))

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\frac{2}{3} log (x (x - 3))

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\frac{2}{3} log (\sqrt{x} (x - 3))

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Solution

The correct option is D $\frac{2}{3} log (\sqrt{x} (x - 3))$
$\int (\begin{matrix} \frac{1}{x - 3} - \frac{1}{x (x - 3)} \end{matrix}) d x = \int (\begin{matrix} \frac{1}{x - 3} + \frac{1}{3 x} - \frac{1}{3 (x - 3)} \end{matrix}) d x = \int (\begin{matrix} \frac{1}{3 x} + \frac{2}{3 (x - 3)} \end{matrix}) d x$

$= \frac{1}{3} log x + \frac{2}{3} log (x - 3) + C$

$= \frac{2}{3} log \sqrt{x} + \frac{2}{3} log (x - 3) + C$

$= \frac{2}{3} log (\sqrt{x} (x - 3)) + C$

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Similar questions

\int (\frac{1}{x - 3} - \frac{1}{x^{2} - 3 x}) d x =

____________

+ c; x > 3

Q. If

\int \frac{x^{2} - x - 1}{\sqrt[3]{1 - 3 x}} d x = \frac{- (x^{2} - x - 1) {(1 - 3 x)}^{\frac{2}{3}}}{2} - \frac{(2 x - 1) {(1 - 3 x)}^{\frac{8}{3}}}{k} + C

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

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Show that $\int \frac{2 x + 3}{x^{2} + 3 x} d x = l o g ∣ ∣ x^{2} + 3 x ∣ ∣ + C$

\frac{x^{2} - x - 3}{x^{3} - 1} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + x + 1} \Rightarrow 2 A + B + C =

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∫(1x−3−1x2−3x) dx= ____ +C, x>3

The correct option is D 23log(√x(x−3))∫(1x−3−1x(x−3))dx=∫(1x−3+13x−13(x−3)) dx=∫(13x+23(x−3))dx =13logx+23log(x−3)+C =23log√x+23log(x−3)+C =23log(√x(x−3))+C

$\int (\begin{matrix} \frac{1}{x - 3} - \frac{1}{x^{2} - 3 x} \end{matrix}) d x =$ ____ $+ C, x > 3$