∫(√(x^2 +4x+13))dx=??


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Question Number 128575 by Study last updated on 08/Jan/21

$\int \sqrt{x^{2} + 4 x + 13} d x = ? ?$
Commented by MJS_new last updated on 08/Jan/21

$use formula for \int \sqrt{x^{2} + a x + b} d x!!! to get$ $\frac{(x + 2) \sqrt{x^{2} + 4 x + 13}}{2} + \frac{9}{2} \ln (x + 2 + \sqrt{x^{2} + 4 x + 13}) + C$
Commented by Study last updated on 08/Jan/21

$w h i c h f o r m u l a ?$
Commented by MJS_new last updated on 08/Jan/21

$you could search the web for integration$ $formulas, {couldn}^{'} t you ?$
	Answered by john_santu last updated on 08/Jan/21

	$\int \sqrt{{(x + 2)}^{2} + 3^{2}} dx$ $x + 2 = 3 \tan t$ $\int \sqrt{9 \sec^{2} t} {3 \sec}^{2} t dt =$ $I = \int 9 \sec^{3} t dt$ $♭ = \int \sec^{3} t dt = \int \sec t d (\tan t)$ $♭ = \sec t . \tan t - \int \tan^{2} t \sec t dt$ $♭ = \sec t . \tan t - \int (\sec^{2} t - 1) \sec t dt$ $♭ = \sec t . \tan t - \int \sec^{3} t dx + \ln ∣ \sec t + \tan t ∣ + c$ $2 ♭ = \sec t . \tan t + \ln ∣ \sec t + \tan t ∣ + c$ $♭ = \frac{\sec t . \tan t}{2} + \frac{\ln ∣ \sec t + \tan t ∣}{2} + C$ $I = \frac{9}{2} \sec t . \tan t + \frac{9}{2} \ln ∣ \sec t + \tan t ∣ + C$ $put \tan t = \frac{x + 2}{3}$
	Answered by MJS_new last updated on 08/Jan/21

	$sometimes you seem like a kid with a bicycle$ $asking how to drive a racing car through a$ $chicane . . .$ $\int \sqrt{x^{2} + 4 x + 13} d x = \int \sqrt{{(x + 2)}^{2} + 9} d x =$ $[t = x + 2 \Leftrightarrow x = t - 2 \to d x = d t]$ $= \int \sqrt{t^{2} + 9} d t =$ $[u = \frac{t + \sqrt{t^{2} + 9}}{3} \Leftrightarrow t = \frac{3 (u^{2} - 1)}{2} \to d t = \frac{3 \sqrt{t^{2} + 9} d u}{t + \sqrt{t^{2} + 9}}]$ $= \frac{9}{4} \int \frac{{(u^{2} + 1)}^{2}}{u^{3}} d u = \frac{9}{4} \int u + \frac{2}{u} + \frac{1}{u^{3}} d u$ $and I hope you can go on from here$
	Commented by john_santu last updated on 08/Jan/21
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	Answered by Dwaipayan Shikari last updated on 08/Jan/21

	$\int \sqrt{{(x + 2)}^{2} + 3^{2}} d x x + 2 = 3 s i n h t \Rightarrow 1 = 3 c o s h t \frac{d t}{d x}$ $= 9 \int {c o s h}^{2} t d t = \frac{9}{2} + \frac{9}{2} \int c o s h (2 t) d t = \frac{9}{2} t + \frac{9}{4} s i n h (2 t) + C$ $= \frac{9}{2} {s i n h}^{- 1} (\frac{x + 2}{3}) + \frac{1}{2} (x + 2) \sqrt{x^{2} + 4 x + 13}$ $= \frac{9}{2} l o g (x + 2 + \sqrt{x^{2} + 4 x + 13}) + \frac{(x + 2)}{2} \sqrt{x^{2} + 4 x + 13} + C$
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