calculus II integration?

Question

evaluate the integral of [ dx /x*sqrt(3+4x ] = integral of "dx" divided by x multiplied by the square root of "3+4x".

answer choices:

a)
-(1/sqrt(3)) ln [ sqrt(3 + 4x) + sqrt(3) / sqrt ( 3+4x) - sqrt(3)]+C

b) (1/sqrt(3)) ln [ sqrt(3+4x) + sqrt(3) / sqrt(3+4x) - sqrt(3) ]+C

c) (1/sqrt(3)) ln [ sqrt(3+4x) - sqrt(3) / sqrt(3+4x) + sqrt(3) ]+ C

d) -(1/sqrt(3)) ln [ sqrt(3+4x) - sqrt(3) / sqrt(3+4x) + sqrt(3)] + C


if you can figure this problem out and explain how you got it i will buy you lunch, i think this problem is done by trig substitution by my answer never comes out exactly how the multiple choice has it.

Answer 1

Try a combination of u substitution and partial fractions.

∫dx/[ x√(3+4x) ]

Now you could let u be sqrt(3+4x) but you'd need to solve for x as well, and the derivative is not that neat.... so try u^2 instead, which makes it easier to find dx and x

u^2 = 3+4x

Now differentiate to find dx in terms of du
2*u du = 4 dx
dx = (1/2) du

And since we need to replace x as well solve u^2 = 3+4x for x x= (u^2 - 3)/ 4

Making these substitutions we get

(1/2)∫u du/[ ((u^2-3)/4 u ]

Simplifying
2∫du/[(u^2-3) ]

Factoring
2∫du/[(u-√3)(u+√3) ]

So we can use partial fractions or consult a table since this form appears on most tables. Partial fractions:
1/(u-√3)(u+√3) = A/(u-√3) + B/(u+√3)
Au + √(3)A + Bu- √(3) B = 1

u coefficient on left is 0 so that gives us
A+B =0
B = -A

constant term on left is 1 so that gives us
√(3)A - √(3)B = 1
2√(3) A = 1
A = 1/[(2√(3)]
B = - 1/[2√(3)]

2∫du/[(u-√3)(u+√3) ] = 1/[(2√(3))u-√3)] - 1/[2√(3) (u+√3)]

Simplify slighlyt by bringing out the constants
1/√(3) ∫du/[(u-√3)(u+√3) ] = 1/√(3) ∫[1/[u-√3)] - 1/(u+√3)] ] du

Finally integrate since they in the for dw/w (if you do another substitution)
1/√(3) * [ln | u-√(3) | - ln(u+√(3))] + C

Combine using the property of logs ln(a) - ln(b) = ln(a/b)
1/√(3) * ln | [u-√(3) ] / [u+√(3) ] + C

Put it back in terms of x
1/√(3) * ln | [√(3+4x)-√(3) ] / [√(3+4x)+√(3) ] +C

which is answer c

Answer 2

♠ t=√(3+4x), t^2 =3+4x, x=0.25(t^2-3), dx=0.5t*dt;
♣ y(t)*dt = 0.5t*dt /(0.25(t^2-3) *t) =2dt /((t-√3) (t+√3))=
= a*dt/(t-√3) + b*dt/(t+√3) =
= dt*[a(t+√3) +b(t-√3)] /((t-√3) (t+√3));
♦ 2 = at +bt +√3(a-b), hence
0=a+b, 2=(a-b)√3; a=1/√3, b=-1/√3;
♥ thus y(t)*dt = dt/(t√3 -3) -dt/(t√3+3);
Y(t) = ln((t√3 -3)) /√3 - ln((t√3 +3)) /√3 =
= ln((t√3 -3)/(t√3 +3)) /√3 =
= ln((√3√(3+4x) -3)/(√3√(3+4x) +3)) /√3 =
= ln((√(3+4x) -√3)/(√(3+4x) +√3)) /√3 +C; (c);