The integral dx (x + 1) ^ 3 4 (x - 2) ^ 5 4 is equal to - Vidyadip

MCQ

The integral $\int {\frac{{dx}}{{{{(x + 1)}^{\frac{3}{4}}}{{(x - 2)}^{\frac{5}{4}}}}}} $ is equal to

A
$ - \frac{4}{3}{\left( {\frac{{x + 1}}{{x - 2}}} \right)^{\frac{1}{4}}}\, + \,c$
✓
$4{\left( {\frac{{x + 1}}{{x - 2}}} \right)^{\frac{1}{4}}}\, + \,c$
C
$4{\left( {\frac{{x - 2}}{{x + 1}}} \right)^{\frac{1}{4}}}\, + \,c$
D
$ - \frac{4}{3}{\left( {\frac{{x - 2}}{{x + 1}}} \right)^{\frac{1}{4}}}\, + \,c$

✓

Answer

Correct option: B.

$4{\left( {\frac{{x + 1}}{{x - 2}}} \right)^{\frac{1}{4}}}\, + \,c$

b
$\int \frac{d x}{(x+1)^{3 / 4}(x-2)^{5 / 4}}$

$\int {\frac{{dx}}{{{{\left( {\begin{array}{*{20}{c}}
{x + 1}\\
{x - 2}
\end{array}} \right)}^{3/4}}{{(x - 2)}^2}}}} $

$\text { put } \frac{x+1}{x-2}=t$

$\frac{{ - 3}}{{{{(x - 2)}^2}}} = \frac{{dt}}{{dx}}$

${\frac{d x}{(x-2)^{2}}=-\frac{d t}{3}} $

$ = \frac{{ - 1}}{3}\int {\frac{{dt}}{{{t^{3/4}}}}} = - \frac{1}{3}\int t \frac{{ - 3}}{4}It$

$ = \frac{1}{3}\left[ {\frac{{{t^{\frac{{ - 3}}{4} + 1}}}}{{\frac{{ - 3}}{4} + 1}}} \right]$

$ = \frac{{ - 4}}{3}{\left[ {\begin{array}{*{20}{c}}
{x + 1}\\
{x - 2}
\end{array}} \right]^{1/4}} + c$

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