Having troubles........

1a)

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

the first thing I notice is this

\(\frac{d}{dx}\;(x^2+4)^{1.5}=1.5(x^2+4)^{0.5}*2x=3x(x^2+4)^{0.5}\\ so\\ \frac{d}{dx}\;\frac{(x^2+4)^{1.5}}{3}=x(x^2+4)^{0.5}\\ so\\ -\int x\sqrt{x^2+4}\; dx = \frac{-(x^2+4)^{1.5}}{3}+c\)

1-a:

Take the integral: - integral x sqrt(x^2+4) dx For the integrand x sqrt(x^2+4), substitute u = x^2+4 and du = 2 x dx: = -1/2 integral sqrt(u) du The integral of sqrt(u) is (2 u^(3/2))/3: = -u^(3/2)/3+constant Substitute back for u = x^2+4: Answer: | | = -1/3 (x^2+4)^(3/2)+constant

1-b:

Take the integral:
- integral e^t sqrt(e^(2 t)+4) dt
For the integrand e^t sqrt(e^(2 t)+4), substitute u = e^t and du = e^t dt:
 = - integral sqrt(u^2+4) du
For the integrand sqrt(u^2+4), substitute u = 2 tan(s) and du = 2 sec^2(s) ds. Then sqrt(u^2+4) = sqrt(4 tan^2(s)+4) = 2 sec(s) and s = tan^(-1)(u/2):
 = -2 integral2 sec^3(s) ds
Factor out constants:
 = -4 integral sec^3(s) ds
Use the reduction formula, integral sec^m(s) ds = (sin(s) sec^(m-1)(s))/(m-1) + (m-2)/(m-1) integral sec^(-2+m)(s) ds, where m = 3:
 = -2 tan(s) sec(s)-2 integral sec(s) ds
The integral of sec(s) is log(tan(s)+sec(s)):
 = -2 tan(s) sec(s)-2 log(tan(s)+sec(s))+constant
Substitute back for s = tan^(-1)(u/2):
 = -2 tan(tan^(-1)(u/2)) sec(tan^(-1)(u/2))-2 log(tan(tan^(-1)(u/2))+sec(tan^(-1)(u/2)))+constant
Simplify using sec(tan^(-1)(z)) = sqrt(z^2+1) and tan(tan^(-1)(z)) = z:
 = -1/2 u sqrt(u^2+4)-2 log(sqrt(u^2+4)+u)+log(4)+constant
Substitute back for u = e^t:
 = -1/2 e^t sqrt(e^(2 t)+4)-2 log(sqrt(e^(2 t)+4)+e^t)+log(4)+constant
Which is equivalent for restricted t values to:
Answer: |= -1/2 e^t sqrt(e^(2 t)+4)-2 sinh^(-1)(e^t/2)+constant

2)

Take the integral:
 integral sqrt(t^2-5 t) dt
For the integrand sqrt(t^2-5 t), complete the square:
 = integral sqrt((t-5/2)^2-25/4) dt
For the integrand sqrt((t-5/2)^2-25/4), substitute u = t-5/2 and du = dt:
 = integral sqrt(u^2-25/4) du
For the integrand sqrt(u^2-25/4), substitute u = (5 sec(s))/2 and du = 5/2 tan(s) sec(s) ds. Then sqrt(u^2-25/4) = sqrt((25 sec^2(s))/4-25/4) = (5 tan(s))/2 and s = sec^(-1)((2 u)/5):
 = 5/2 integral5/2 tan^2(s) sec(s) ds
Factor out constants:
 = 25/4 integral tan^2(s) sec(s) ds
For the integrand tan^2(s) sec(s), write tan^2(s) as sec^2(s)-1:
 = 25/4 integral sec(s) (sec^2(s)-1) ds
Expanding the integrand sec(s) (sec^2(s)-1) gives sec^3(s)-sec(s):
 = 25/4 integral(sec^3(s)-sec(s)) ds
Integrate the sum term by term and factor out constants:
 = 25/4 integral sec^3(s) ds-25/4 integral sec(s) ds
Use the reduction formula, integral sec^m(s) ds = (sin(s) sec^(m-1)(s))/(m-1) + (m-2)/(m-1) integral sec^(-2+m)(s) ds, where m = 3:
 = 25/8 tan(s) sec(s)-25/8 integral sec(s) ds
The integral of sec(s) is log(tan(s)+sec(s)):
 = 25/8 tan(s) sec(s)-25/8 log(tan(s)+sec(s))+constant
Substitute back for s = sec^(-1)((2 u)/5):
 = 25/8 sec(sec^(-1)((2 u)/5)) tan(sec^(-1)((2 u)/5))-25/8 log(sec(sec^(-1)((2 u)/5))+tan(sec^(-1)((2 u)/5)))+constant
Simplify using sec(sec^(-1)(z)) = z and tan(sec^(-1)(z)) = sqrt(1-1/z^2) z:
 = 1/4 u sqrt(4 u^2-25)-25/8 log(1/5 (sqrt(4 u^2-25)+2 u))+constant
Substitute back for u = t-5/2:
 = 1/4 sqrt((t-5) t) (2 t-5)-25/8 log((2 t)/5+2/5 sqrt((t-5) t)-1)+constant
Factor the answer a different way:
 = 1/8 (2 sqrt((t-5) t) (2 t-5)-25 log((2 t)/5+2/5 sqrt((t-5) t)-1))+constant
Which is equivalent for restricted t values to:
Answer: |= (t (2 t^2-15 t+25)-25 sqrt(t-5) sqrt(t) log(sqrt(t-5)+sqrt(t)))/(4 sqrt((t-5) t))+constant

Thank you both :)