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

Question Number 112251 by bobhans last updated on 07/Sep/20

\int \frac{dx}{(x^{4} - 1) \sqrt{x^{2} + 1}}

Answered by john santu last updated on 07/Sep/20

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

s e t t i n g : \tan r = x \Rightarrow d x = \sec^{2} r d r

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

I = \int \frac{\sec^{2} r d r}{(\tan^{2} r - 1) \sec^{3} r}

I = \int \frac{\cos r d r}{\tan^{2} r - 1} = \int \frac{\cos^{3} r d r}{\sin^{2} r - \cos^{2} r}

I = \int \frac{(1 - \sin^{2} r) \cos r d r}{2 \sin^{2} r - 1}

l e t \sin r = q \Rightarrow I = \int \frac{1 - q^{2}}{2 q^{2} - 1} d q

I = \frac{1}{2} \int (\frac{1}{2 q^{2} - 1} - 1) d q

I = \frac{1}{4} \int (- 2 + \frac{1}{q \sqrt{2} - 1} - \frac{1}{q \sqrt{2} + 1}) d q

I = \frac{1}{4} (- 2 q + \frac{1}{\sqrt{2}} \ln ∣ q \sqrt{2} - 1 ∣ - \frac{1}{\sqrt{2}} \ln ∣ q \sqrt{2} + 1 ∣) + c

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

Commented by bemath last updated on 07/Sep/20

s a n t u y y y

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