First one:
$\displaystyle \displaystyle \begin{align*} \int{ \left( \frac{4.8}{1 + 6e^{-0.46x}} \right)^2 \,dx} &= \int{ \frac{ -\frac{50}{23} e^{ -\frac{23}{50} x } }{ -\frac{50}{23} e^{-\frac{23}{50} x} } \left( \frac{4.8}{1 + 6e^{ -\frac{23}{50} x} } \right)^2 \,dx} \end{align*}$
Now make the substitution $\displaystyle \displaystyle \begin{align*} u = e^{-\frac{23}{50}x} \implies du = -\frac{50}{23}e^{-\frac{23}{50}x}\,dx \end{align*}$ and the integrate using Partial Fractions.
For the second problem, expand the integrand, then integrate term by term.
The exponential terms may be integrated by beginning with the derivative:
$\displaystyle \frac{d}{dx}\left(4^{kx} \right)=k\ln(4)4^{kx}$ hence:
$\displaystyle \frac{1}{k\ln(4)}\int\,d\left(4^{kx} \right)=\int 4^{kx}\,dx$
$\displaystyle \int 4^{kx}\,dx=\frac{4^{kx}}{k\ln(4)}+C$
Thank you again but now im sort of getting embarassed because I never learnt any partial fractions, so even though I looked at series of websites, that explains partial fraction, I still can't apply it... I know it's really annoying, but could you help me just a little bit more???
You begin by assuming the integrand may be expressed in the form:
$\displaystyle \frac{1}{u(6u+1)^2}=\frac{A}{u}+\frac{B}{6u+1}+ \frac{C}{(6u+1)^2}$
Now multiply through by the lowest common denominator $\displaystyle u(6u+1)^2$
$\displaystyle 1=A(6u+1)^2+Bu(6u+1)+Cu$
Now, expand the right side and arrange as a quadratic in standard form, and write the left side as:
$\displaystyle 0u^2+0u+1$
Then equate corresponding coefficients to arrive at a linear 3X3 system in A, B, and C for which there is a unique solution.
Thank you for your help once again, but would it be asking too much if I ask you to solve this problem (the second one) for me? I know you are only meant to give directions and helps but I have a very limited knowledge about partial fractions. Also, I have like 5 of questions that look similar but with completely different numbers, and other 10-15 questions completely different. So could you please do me a favour just once?
Totally understood. What I have done so far is from your advice and MarkFL2's advice. what I have got is
1=36Au^2+12Au+A+6Bu^2+Bu+Cu
which is basically an expansion of 1=A(6u+1)^2+Bu(6u+1)+Cu.
I have no idea how to go from here..
You want to arrange both sides of the equation in standard form as quadratics, i.e,:
$\displaystyle 0u^2+0u+1=(36A+6B)u^2+(12A+B+C)u+A$
Now, equating the corresponding coefficients on both sides, will give you 3 equations, from which you may determine A, B and C.
You have two of the values correct. Did you try substituting the solutions into the equations to make sure all 3 equations are satisfied?
Once you have the correct solutions, then you may put their values into:
$\displaystyle \frac{1}{u(6u+1)^2}=\frac{A}{u}+\frac{B}{6u+1}+ \frac{C}{(6u+1)^2}$
to get the partial fraction decomposition of the original integral. You will then have 3 terms which you can easily integrate.