What are the Isomers of Pentane?

Pentane (C5H12) has three structural isomers. Isomer 1 is n-pentane, the straight chain normal structure for pentane Isomer 2 is 2-methylbutane, a branched chain with a carbon* atom joined onto three other carbon atoms. Isomer 3 is 2,2-dimethylpropane, a branched chain with the central carbon* atom joined onto four other carbon atoms. Note that for all the isomers, each carbon atom has four bonds (valency 4), and each hydrogen atom has one bond (valency 1). Valency is the combining power of an atom. You cannot change the number of bonds that each atom has.

How do you evaluate the integral ∫ x^4/(1+x^2) dx ?

x*4/(x*2+1) = (x^2-1) + 1/1+x^2 So, integrating both sides , we get x^3/3 - x + ∫dx/1+x^2 now let x = tanθ , dx = sec^2θdθ and x^2 + 1 = 1 + tan^2θ = sec^2θ therefore, ∫dx/1+x^2 = ∫dθ = θ = arctan(x) putting it all together ∫x*4/(x*2+1) = x^3/3 - x +arctan(x) + C

Find the roots of f(x) = 2x3 + 3x2 – 11x – 6 = 0, given that it has at least one integer root.

ince the constant in the given equation is a 6, we know that the integer root must be a factor of 6. The possible values are Step 1: Use the factor theorem to test the possible values by trial and error. f(1) = 2 + 3 – 11 – 6 ≠ 0 f(–1) = –2 + 3 + 11 – 6 ≠ 0 f(2) = 16 + 12 – 22 – 6 = 0 We find that the integer root is 2. Step 2: Find the other roots either by inspection or by synthetic division. 2x3 + 3x2 – 11x – 6 = (x – 2)(ax2 + bx + c) = (x – 2)(2x2 + bx + 3) = (x – 2)(2x2 + 7x + 3) = (x – 2)(2x + 1)(x +3) So, the roots are x=2,-0.5,-3