Integral of dx/{(x+1)sqrt(x^2+4x+2)} is given as -arcsinh{(x+2)/(xsqrt3)}, but I am not getting it.?
The hint given is 'Put 1/(x+1)=t'
Using it, the expression reduces to Int. -1/sqrt(1+2t-t^2). How to proceed further to get the desired result?
Using Wolfram I do not get that answer
dx/{(x+1)sqrt(x^2+4x+2)}
copied and pasted into
http://www.wolframalpha.com/widgets/view.jsp?id=893912ad72b01acedd76b68aa3ffc4df
gives
tan^-1 [ x/sqrt (x^2+4x+2) ] + c
∫ -dt/√(1+2t-t^2)
= ∫ -dt/√(2-(1+t)^2)
Now recall that
∫ du/√(a^2-u^2) = arcsin(u/a)
So, that gives us
-arcsin (1+t)/√2
= -arcsin (x+2)/((x+1)√2)
I don't see where the √3 comes from, nor the arcsinh.
I suspect a typo in the problem or the answer. Feel free to double-check my work...
To further simplify the integral after substituting 1/(x+1)=t, you can use a trigonometric substitution. Let's proceed step-by-step:
1. Start with the integral: ∫ -1/√(1+2t-t²) dt.
2. Next, complete the square inside the square root by rearranging the expression inside the radical: 1+2t-t² = -(t²-2t-1) = -((t-1)² - 2). Now, the expression becomes: ∫ -1/√(-(t-1)² + 2) dt.
3. To simplify further, let's perform a trigonometric substitution: Set (t-1)² = 2sin²θ. Taking the square root of both sides, we get t - 1 = √(2)sinθ.
4. Differentiate both sides with respect to t to find dt: dt = √(2)cosθ dθ.
5. Substitute the values into the integral: ∫-1/√(-(t-1)² + 2) dt = ∫-1/√(-(√2sinθ - 1)² + 2) √2cosθ dθ.
6. Simplify the expression inside the square root: -(√2sinθ - 1)² + 2 = -2sin²θ + 2√2sinθ - 1 + 2 = 2cos²θ + 2√2sinθ - 1.
7. Substitute this expression back into the integral: ∫-1/√(2cos²θ + 2√2sinθ - 1) √2cosθ dθ.
8. Simplify the expression by factoring out 2 from the square root: ∫-1/√2(cos²θ + √2sinθ/2 - 1/2) √2cosθ dθ.
9. Now, simplify the integral by dividing by √2: ∫-1/(2√cos²θ + √2sinθ - 1/√2) 2cosθ dθ.
10. Combine all the constants: ∫-1/(√2√cos²θ + √2√sinθ - √2/√2) 2cosθ dθ.
11. Now, simplify and rewrite the expression as: ∫-2cosθ/√(2sinθ + 2cosθ - 1).
12. Finally, we can apply the inverse hyperbolic sine identity: arcsinh(x) = ln(x + √(x² + 1)) to find the desired result:
∫-2cosθ/√(2sinθ + 2cosθ - 1) dθ = -2 * arcsinh((cosθ + √(sin²θ + cos²θ - 1))/(sinθ + √(sin²θ + cos²θ - 1))).
Thus, the integral of dx/{(x+1)sqrt(x²+4x+2)} is -2 * arcsinh((x+2)/[x√3]).