Everyday Math 每日數學 之 Integration #20190517 (+Solution)

(a)
$ e^x=\frac{1}{t-1} $
$ x=-\ln(t-1) $
$ dx=-\frac{1}{t-1}dt $
$ \int \frac{1}{e^{4x}+e^{3x}}dx $
$ =\int \frac{1}{\left(\frac{1}{t-1}\right)^{4}+\left(\frac{1}{t-1}\right)^{3}}\cdot -\frac{1}{t-1}dt $
$ =-\int \frac{1}{\left(\frac{1}{t-1}\right)^{3}+\left(\frac{1}{t-1}\right)^{2}} dt $
$ =-\int \frac{(t-1)^3}{1+t-1} dt $
$ =-\int \frac{t^3-3t^2+3t-1}{t} dt $
$ =-\frac{1}{3}t^3+\frac32t^2-3t+\ln|t|+C $
$ =-\frac{1}{3}\left( \frac{1}{e^x}+1\right)^3+\frac32\left( \frac{1}{e^x}+1\right)^2-3\left( \frac{1}{e^x}+1\right)+\ln\left| \frac{1}{e^x}+1\right|+C $

(b)
Let $ u=e^x $ . Then $ du=e^x dx $ .
$ \int \frac{1}{u^5+u^4}du $
$ =\int \frac{1}{e^{5x}+e^{4x}}\cdot e^xdx $
$ =\int \frac{1}{e^{4x}+e^{3x}}dx $
$ =-\frac{1}{3}\left( \frac{1}{e^x}+1\right)^3+\frac32\left( \frac{1}{e^x}+1\right)^2-3\left( \frac{1}{e^x}+1\right)+\ln\left| \frac{1}{e^x}+1\right|+C $ ( by (a) )
$ =-\frac{1}{3}\left( \frac{1}{u}+1\right)^3+\frac32\left( \frac{1}{u}+1\right)^2-3\left( \frac{1}{u}+1\right)+\ln\left|\frac{1}{u}+1\right|+C $

(c)
$ \int _1^e \frac{5u^5+4u^4}{(u^5+u^4)^2}du $
$ =\int _1^e \frac{(5u^4+4u^3)u}{(u^5+u^4)^2}du $
$ =\int _{u=1}^{u=e} \frac{u}{(u^5+u^4)^2}d(u^5+u^4) $
$ =\int _{u=1}^{u=e} u(u^5+u^4)^{-2}d(u^5+u^4) $
$ =-\int _{u=1}^{u=e} ud(u^5+u^4)^{-1} $
$ =- \left[ u(u^5+u^4)^{-1}\right]_{u=1}^{u=e} +\int _{u=1}^{u=e} (u^5+u^4)^{-1}du $
$ =- \left[ (u^4+u^3)^{-1}\right]_{u=1}^{u=e} +\left[-\frac{1}{3}\left( \frac{1}{u}+1\right)^3+\frac32\left( \frac{1}{u}+1\right)^2-3\left( \frac{1}{u}+1\right)+\ln\left| \frac{1}{u}+1\right|\right]_{u=1}^{u=e} $
$ =- (e^4+e^3)^{-1} -\frac{1}{3}\left( \frac{1}{e}+1\right)^3+\frac32\left( \frac{1}{e}+1\right)^2-3\left( \frac{1}{e}+1\right)+\ln\left( \frac{1}{e}+1\right)+\frac{19}{6}-\ln 2 $