Evaluate-x-2x-2-4x-7-dx-

Question Number 179511 by Acem last updated on 30/Oct/22

E v a l u a t e \int \frac{x}{\sqrt{2 x^{2} - 4 x - 7}} d x

Answered by ARUNG_Brandon_MBU last updated on 30/Oct/22

= \frac{1}{4} \int \frac{4 x - 4}{\sqrt{2 x^{2} - 4 x - 7}} d x + \int \frac{d x}{\sqrt{2 {(x - 1)}^{2} - 9}}

= \frac{\sqrt{2 x^{2} - 4 x - 7}}{2} + \frac{\sqrt{2}}{2} \arcsin (\frac{\sqrt{2} (x - 1)}{3}) + C

Commented by Acem last updated on 30/Oct/22

T h a n k s!

Commented by Ar Brandon last updated on 30/Oct/22

You're welcome!

Commented by Frix last updated on 31/Oct/22

\int \frac{d x}{\sqrt{2 {(x - 1)}^{2} - 9}} = \frac{\sqrt{2}}{2} \cosh^{- 1} \frac{\sqrt{2} (x - 1)}{3}

Commented by Ar Brandon last updated on 08/Nov/22

You're right. Thanks

Answered by Acem last updated on 30/Oct/22

\int \frac{x}{\sqrt{2 x^{2} - 4 x - 7}} d x

2 x^{2} - 4 x - 7 = 2 (x^{2} - 2 x + 1 - \frac{9}{2}) = 2 {(x - 1)}^{2} - 9

x - 1 = \frac{3}{\sqrt{2}} \sec θ, d x = \frac{3}{\sqrt{2}} \sec θ \tan θ d θ

\sqrt{2 x^{2} - 4 x - 7} = 3 \sqrt{\sec^{2} θ - 1} = 3 ∣ \tan θ ∣= 3 \tan θ

I = \frac{1}{\sqrt{2}} \int (1 + \frac{3}{\sqrt{2}} \sec θ) \sec θ d θ

I = \frac{1}{\sqrt{2}} \ln ∣ \sec θ + \tan θ ∣ + \frac{3}{2} \tan θ + c

S i n c e \sec θ = \frac{\sqrt{2} (x - 1)}{3} \Rightarrow \tan θ = \frac{\sqrt{2 {(x - 1)}^{2} - 9}}{3}

I = \frac{1}{\sqrt{2}} \ln ∣ \frac{\sqrt{2} (x - 1) + \sqrt{2 x^{2} - 4 x - 7}}{3} ∣ + \frac{1}{2} \sqrt{2 x^{2} - 4 x - 7} + c

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