The correct option is B (x^2+2x+5)^3/2 (x+1)/2√(x^2+2x+5) 2sinh^ 1(x+1/2)+c I=∫( 3x+2 ) #160; √( x ^ 2 +2x+5 ) dx Now 3x+2 can be written as ...

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Trigonometric Equations

∫3x+2√x2+2x+5...

Question

$\int (3 x + 2) \sqrt{(x^{2} + 2 x + 5)} d x =$

(x^{2} + 2 x + 5)^{3 / 2} - \frac{(x - 1)}{2} \sqrt{x^{2} + 2 x + 5} - 2 {sinh}^{- 1} (\frac{x + 1}{2}) + c

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(x^{2} + 2 x + 5)^{3 / 2} - \frac{(x + 1)}{2} \sqrt{x^{2} + 2 x + 5} - 2 {sinh}^{- 1} (\frac{x + 1}{2}) + c

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(x^{2} - 2 x - 5)^{3 / 2} - \frac{(x - 1)}{2} \sqrt{x^{2} + 2 x + 5} - 2 {sinh}^{- 1} (\frac{x + 1}{2}) + c

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(x^{2} + 2 x + 5)^{3 / 2} - \frac{(x - 1)}{2} \sqrt{x^{2} - 2 x - 5} - 2 {sinh}^{- 1} (\frac{x - 1}{2}) + c

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Solution

The correct option is B $(x^{2} + 2 x + 5)^{3 / 2} - \frac{(x + 1)}{2} \sqrt{x^{2} + 2 x + 5} - 2 {sinh}^{- 1} (\frac{x + 1}{2}) + c$
$I = \int (3 x + 2) \sqrt{x^{2} + 2 x + 5} d x$
Now $3 x + 2$ can be written as $\frac{3}{2} [2 x + 2] - 1$
$∴ I = \int \frac{3}{2} [(2 x + 2) \sqrt{x^{2} + 2 x + 5}] d x - \int \sqrt{x^{2} + 2 x + 5} d x$
$\Rightarrow I = A - B . . . . . (i)$
Where $A = \int \frac{3}{2} (2 x + 2) \sqrt{x^{2} + 2 x + 5} d x$
Put $x^{2} + 2 x + 5 = z$ $\Rightarrow 2 x + 2 = \frac{d z}{d x}$
i.e. $(2 x + 2) d x = d z$
$∴ A = \int \frac{3}{2} \sqrt{z} d z = \frac{3}{2} \frac{z^{3 / 2}}{3 / 2} = z^{3 / 2}$
$A = {(x^{2} + 2 x + 5)}^{3 / 2}$
$B = \int \sqrt{x^{2} + 2 x + 5} d x = \int \sqrt{{(x + 1)}^{2} + 2^{2}} d x$
Now, $\int \sqrt{x^{2} + a^{2}} d x = \frac{x \sqrt{x^{2} + a^{2}}}{2} + \frac{a^{2}}{2} {sinh}^{- 1} \frac{x}{a}$
$∴ B = \frac{(x + 1)}{2} \sqrt{x^{2} + 2 x + 5} + \frac{2^{2}}{2} {sinh}^{- 1} (\frac{x + 1}{2}) + C$
Substituting $A$ and $B$ in equation $(i)$ , we get
$I = {(x^{2} + 2 x + 5)}^{3 / 2} - \frac{(x + 1)}{2} \sqrt{x^{2} + 2 x + 5} - 2 {sinh}^{- 1} (\frac{x + 1}{2}) + C$

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