Show that $\int \Big(\frac{98}{1+19e^{-2x}} -4.9\ \Big)dx=49\ln(e^{2x}+19)-4.9x+c$ As per the question $$\int \Big(\frac{98}{1+19e^{-2x}} -4.9\ \Big)dx=49\ln(e^{2x}+19)-4.9x+c$$
I'm unsure how to step through this question. Can anyone please help?
 A: we have to solve $$\int\frac{e^{2x}}{e^{2x}+19}dx$$ setting $$e^{2x}=t$$ then we get
$$e^{2x}dx=\frac{1}{2}dt$$ and our integr5al will be $$\frac{1}{2}\int\frac{dt}{t+19}$$
A: Well, this equation is easier to solve from the right hand side, as the differentiation is much easier.
Take the derivatives of r.h.s w.r.t x:
$$
\nabla_x rhs = \nabla_x 49\ln(e^{2x}+19)-4.9x+c \\
= 49\frac{1}{e^{2x} + 19}e^{2x}\times 2 - 4.9 \\
= 98\frac{1}{1+19e^{-2x}} -4.9  \\
= \frac{98}{1+19e^{-2x}} -4.9 \\
= lhs
$$
Inspect the process in a backward manner, you shall get the point.
Of course, the $\int\frac{98}{1+19e^{-2x}} dx$ can be integrated by substitution.
Let:
$$
t=e^{-2x}
$$
Then:
$$
\int\frac{98}{1+19e^{-2x}} dx\\
= \int \frac{98}{1+19t}d(-\dfrac{1}{2}\ln t) \\
= \int \frac{-49}{1+19t}\frac{19}{19t}dt \\
= \int -49\times 19(\frac{1}{19t} - \frac{1}{1 + 19t})dt \\
= -49\times 19(\frac{1}{19}\ln t - \frac{1}{19}\ln(1 + 19t) + c \\
= -49\ln\frac{t}{t+19} + c \\
= 49\ln(1 + \frac{19}{t}) + c\\
= 49\ln(1 + e^{-2x}) + c
$$
