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$\int \frac{dx}{\left ( 2x+3 \right )\sqrt{4x+5}}=$

Option 1)
$\tan^{-1}\sqrt{4x-5}+c$

Option 2)
$\tan^{-1}\sqrt{4x+5}+c$

Option 3)
$\tan^{-1}\sqrt{5x+4}+c$

Option 4)
$\tan^{-1}\sqrt{5x-4}+c$

Answers (1)

best_answer

As we learnt

Type of Integration by perfect square -

The integration in the form

(i) $\int \frac{dx}{(px+q)\sqrt{ax^{2}+bx+c}}$

(ii) $\int \frac{dx}{(px+q)\sqrt{ax+b}}$

(iii) $\int \frac{(a+bx)^{m}}{(p+qx)^{n}}dx$

- wherein

Working rule.

(i) $\rightarrow$ put $(px+q)=\frac{1}{t}$

(ii) $\rightarrow$ put $(ax+b)=t^{2}$

(iii) $\rightarrow$ put $(p+qx)=t$

$Let I=\int {\frac{{dx}}{{\left( {2x + 3} \right)\sqrt {4x + 5} }}} $$

Put 4x + 5 = t

$\therefore x=\frac{{t - 5}}{4}$$ $\therefore x=\frac{{t - 5}}{4}\: \: \therefore dx= \frac{dt}{4}$

$\therefore I=\frac{1}{4}\int {\frac{{dt}}{{\left( {\frac{{2t - 10}}{4} + 3} \right)\sqrt t }}} =\frac{1}{2}\int {\frac{{dt}}{{(t + 1)\sqrt t }}}$

$Let\: \sqrt{t}=u$

$\therefore I=\int {\frac{{du}}{{{u^2} + 1}}} = {\tan ^{ - 1}}\sqrt t + c$$

$\therefore I=\tan^{-1}\sqrt {4x + 5} + c$ .

Option 1)

$\tan^{-1}\sqrt{4x-5}+c$

Option 2)

$\tan^{-1}\sqrt{4x+5}+c$

Option 3)

$\tan^{-1}\sqrt{5x+4}+c$

Option 4)

$\tan^{-1}\sqrt{5x-4}+c$

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