A calculus problem by Akshat Sharda

Calculus Level 5

∫ d x ( x 2 + 2 x + 2 ) x 2 + 2 x + 4 \int \dfrac{dx}{(x^2+2x+2)\sqrt{x^2+2x+4}}

Find the above integral.

Notation: C C in the answer options denotes the constant of integration .

None of these. 1 2 tan ⁡ − 1 ( 1 x + 1 x 2 + 2 x + 2 2 ) + C \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{1}{x+1} \sqrt{\frac{x^2+2x+2}{2}}\right) + C − 1 2 tan ⁡ − 1 ( 1 x + 1 x 2 + 2 x + 4 2 ) + C -\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{1}{x+1} \sqrt{\frac{x^2+2x+4}{2}}\right) + C − 1 2 tan ⁡ − 1 ( 1 x + 2 x 2 + 2 x + 4 4 ) + C -\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{1}{x+2} \sqrt{\frac{x^2+2x+4}{4}}\right) + C 1 2 tan ⁡ − 1 ( 1 x + 4 x 2 + 2 x + 4 2 ) + C \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{1}{x+4} \sqrt{\frac{x^2+2x+4}{2}}\right) + C

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

∫ d x ( ( x + 1 ) 2 + 1 ) ( x + 1 ) 2 + 3 \large \displaystyle \int _{ }^{ }\frac{dx}{\left(\left(x+1\right)^2+1\right)\sqrt{\left(x+1\right)^2+3}}

Let x + 1 = 3 tan ⁡ t ⟹ d x = 3 sec ⁡ 2 t d t \large \displaystyle x+1 = \sqrt{3}\tan t \implies dx = \sqrt{3}\sec ^2t\ dt

∫ 3 sec ⁡ 2 t d t ( 3 tan ⁡ 2 t + 1 ) 3 sec ⁡ t = ∫ sec ⁡ t d t 3 tan ⁡ 2 t + 1 \large \displaystyle \int _{ }^{ }\frac{\sqrt{3}\sec ^2t\ dt}{\left(3\tan ^2t+1\right)\sqrt{3}\sec t} = \int _{ }^{ }\frac{\sec tdt}{3\tan ^2t+1}

Multiplying top and bottom by cos ⁡ 2 t \large \displaystyle \cos ^2t ,

∫ cos ⁡ t d t 3 sin ⁡ 2 t + cos ⁡ 2 t = ∫ cos ⁡ t d t 3 sin ⁡ 2 t + 1 − sin ⁡ 2 t = ∫ cos ⁡ t d t 2 sin ⁡ 2 t + 1 \large \displaystyle \int _{ }^{ }\frac{\cos tdt}{3\sin ^2t+\cos ^2t} = \int _{ }^{ }\frac{\cos tdt}{3\sin ^2t+1-\sin ^2t} = \int _{ }^{ }\frac{\cos tdt}{2\sin ^2t+1}

Let u = sin ⁡ t ⟹ d u = cos ⁡ t d t \large \displaystyle u = \sin t \implies du = \cos t\ dt . hence,

= ∫ d u 2 u 2 + 1 = ∫ d u ( 2 u ) 2 + 1 = 1 2 tan ⁡ − 1 ( 2 u ) + C \large \displaystyle =\int _{ }^{ }\frac{du}{2u^2+1} = \int _{ }^{ }\frac{du}{\left(\sqrt{2}u\right)^2+1} = \frac{1}{\sqrt{2}}\tan ^{-1}\left(\sqrt{2}u\right)+C

= 1 2 tan ⁡ − 1 ( 2 sin ⁡ t ) + C \large \displaystyle = \frac{1}{\sqrt{2}}\tan ^{-1}\left(\sqrt{2}\sin t\right)+C

= 1 2 tan ⁡ − 1 ( 2 ( x + 1 ) x 2 + 2 x + 4 ) + C \large \displaystyle = \frac{1}{\sqrt{2}}\tan ^{-1}\left(\frac{\sqrt{2}\left(x+1\right)}{\sqrt{x^2+2x+4}}\right)+C

= 1 2 ( π 2 − cot ⁡ − 1 ( 2 ( x + 1 ) x 2 + 2 x + 4 ) ) + C \large \displaystyle = \frac{1}{\sqrt{2}}\left(\frac{\pi }{2}-\cot ^{-1}\left(\frac{\sqrt{2}\left(x+1\right)}{\sqrt{x^2+2x+4}}\right)\right)+C

= π 2 2 − 1 2 cot ⁡ − 1 ( 2 ( x + 1 ) x 2 + 2 x + 4 ) + C \large \displaystyle = \frac{\pi }{2\sqrt{2}}-\frac{1}{\sqrt{2}}\cot ^{-1}\left(\frac{\sqrt{2}\left(x+1\right)}{\sqrt{x^2+2x+4}}\right)+C

= π 2 2 − 1 2 tan ⁡ − 1 ( 1 x + 1 x 2 + 2 x + 4 2 ) + C \large \displaystyle = \frac{\pi }{2\sqrt{2}}-\frac{1}{\sqrt{2}}\tan ^{-1}\left(\frac{1}{x+1}\sqrt{\frac{x^2+2x+4}{2}}\right)+C

= − 1 2 tan ⁡ − 1 ( 1 x + 1 x 2 + 2 x + 4 2 ) + C + π 2 2 \large \displaystyle = -\frac{1}{\sqrt{2}}\tan ^{-1}\left(\frac{1}{x+1}\sqrt{\frac{x^2+2x+4}{2}}\right) + \boxed{C + \frac{\pi }{2\sqrt{2}}}

= − 1 2 tan ⁡ − 1 ( 1 x + 1 x 2 + 2 x + 4 2 ) + C ′ \large \displaystyle = -\frac{1}{\sqrt{2}}\tan ^{-1}\left(\frac{1}{x+1}\sqrt{\frac{x^2+2x+4}{2}}\right) + C'

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