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

Question Number 96175 by Fikret last updated on 30/May/20

\int \frac{d x}{\sqrt{4 x^{2} + 4 x + 3}} = ?

Commented by bobhans last updated on 30/May/20

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

s e t x + \frac{1}{2} = \sqrt{\frac{1}{2}} \tan u \Rightarrow \tan u = \frac{2 x + 1}{\sqrt{2}}

= \frac{1}{2} \int \frac{\frac{1}{\sqrt{2}} \sec^{2} u du}{\frac{1}{\sqrt{2}} \sec u} = \frac{1}{2} \ln ∣ \sec u + \tan u ∣ + c

Answered by mathmax by abdo last updated on 30/May/20

I = \int \frac{dx}{\sqrt{{4 x}^{2} + 4 x + 3}} \Rightarrow I = \int \frac{dx}{\sqrt{{(2 x)}^{2} + 2 (2 x) + 1 + 2}}

= \int \frac{dx}{\sqrt{{(2 x + 1)}^{2} + 2}} =_{2 x + 1 = \sqrt{2} u} \int \frac{du}{\sqrt{2} \sqrt{2} \sqrt{1 + u^{2}}} = \frac{1}{2} argsh (u) + c

= \frac{1}{2} \ln (u + \sqrt{1 + u^{2}}) + c = \frac{1}{2} \ln (\frac{2 x + 1}{\sqrt{2}} + \sqrt{1 + {(\frac{2 x + 1}{\sqrt{2}})}^{2}}) + c