2x-1-x-2-4x-1-3-2-dx-

Question Number 36545 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

\int \frac{2 x + 1}{{(x^{2} + 4 x + 1)}^{\frac{3}{2}}} d x

Commented by MJS last updated on 03/Jun/18

\dots I just gave it a try because the “ {poper}^{″}

way looked rather complicated \dots

Commented by MJS last updated on 03/Jun/18

this looks like {it}^{'} s going to be

\frac{a x + b}{\sqrt{x^{2} + 4 x + 1}}

{(\frac{a x + b}{\sqrt{x^{2} + 4 x + 1}})}^{'} =

Missing \left or extra \right

2 a - b = 2

a - 2 b = 1

a = 1

b = 0

\int \frac{2 x + 1}{{(x^{2} + 4 x + 1)}^{\frac{3}{2}}} d x = \frac{x}{\sqrt{x^{2} + 4 x + 1}} + C

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

e x c e l l e n t s i r \dots

Commented by ajfour last updated on 03/Jun/18

{c o u l d n}^{'} t f o l l o w \dots ? ?

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

s i r m j s a s s u m e d t h e r e s u l t s o f t h e i n t r e g a l

a s \frac{a x + b}{\sqrt{x^{2} + 4 x + 1}} t h e n d i f f e r e n t i a t e d a n d e q u a t e x

a n d g o t t h e r e s u l t s

Commented by ajfour last updated on 03/Jun/18

t h a n k s, i s e e .

Answered by ajfour last updated on 03/Jun/18

= \int \frac{2 x + 1}{{[{(x + 2)}^{2} - {(\sqrt{3})}^{2}]}^{3 / 2}} d x

l e t x + 2 = \sqrt{3} \sec θ

\Rightarrow d x = \sqrt{3} \sec θ \tan θ d θ

I = \int \frac{2 \sqrt{3} \sec θ - 3}{3 \sqrt{3} \sec^{3} θ} d θ

= \frac{2}{3} \int \cos^{2} θ d θ - \frac{1}{\sqrt{3}} \int \cos^{3} θ d θ

= \frac{1}{3} (θ + \frac{\sin 2 θ}{2}) - \frac{1}{4 \sqrt{3}} \int (\cos 3 θ + 3 \cos θ) d θ

= \frac{θ}{3} + \frac{\sin 2 θ}{6} - \frac{\sin 3 θ}{12 \sqrt{3}} - \frac{\sqrt{3}}{4} \sin θ + c .

Answered by MJS last updated on 03/Jun/18

\int \frac{2 x + 1}{{(x^{2} + 4 x + 1)}^{\frac{3}{2}}} d x = \int \frac{2 x + 4}{{(x^{2} + 4 x + 1)}^{\frac{3}{2}}} d x - 3 \int \frac{d x}{{(x^{2} + 4 x + 1)}^{\frac{3}{2}}} =

[\begin{matrix} \int \frac{2 x + 4}{{(x^{2} + 4 x + 1)}^{\frac{3}{2}}} d x = \\ [t = x^{2} + 4 x + 1 \to d x = \frac{d t}{2 x + 4}] \\ = \int \frac{d t}{t^{\frac{3}{2}}} = - \frac{2}{\sqrt{t}} = - \frac{2}{\sqrt{x^{2} + 4 x + 1}} \end{matrix}]

[\begin{matrix} \int \frac{d x}{{(x^{2} + 4 x + 1)}^{\frac{3}{2}}} = \int \frac{d x}{{({(x + 2)}^{2} - 3)}^{\frac{3}{2}}} = \\ [u = x + 2 \to d x = d u] \\ = \int \frac{d u}{{(u^{2} - 3)}^{\frac{3}{2}}} = \\ [v = \sqrt{3} \sec v \to d u = \sqrt{3} \sec v \tan v d v] \\ = \int \frac{\sqrt{3} \sec v \tan v}{{({3 \sec}^{2} v - 3)}^{\frac{3}{2}}} d v = \frac{1}{3} \int \frac{\sec v}{\tan^{2} v} d v = \frac{1}{3} \int \frac{\cos v}{\sin^{2} v} d v = \\ [w = \sin v \to d v = \frac{d w}{\cos v}] \\ = \frac{1}{3} \int \frac{d w}{w^{2}} = - \frac{1}{3 w} = - \frac{1}{3 \sin v} = - \frac{u}{3 \sqrt{u^{2} - 3}} = \\ = - \frac{x + 2}{3 \sqrt{x^{2} + 4 x + 1}} \end{matrix}]

= \frac{x}{\sqrt{x^{2} + 4 x + 1}} + C

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