# Integral of [a(x^n)+b]/[c(x)+d]

\begin{align*}
I = \displaystyle \int_\alpha^\beta
\dfrac{a(x)^n+b}{cx+d} \: dx &=\displaystyle \dfrac{a}{c} \int_\alpha^\beta
\dfrac{(x)^n+ \frac{b}{a}}{x+\frac{d}{c}} \: dx \\&=\displaystyle \dfrac{a}{c} \int_\alpha^\beta
\dfrac{(x)^n+ \Delta_1}{x+\Delta_2} \: dx \\&\therefore \: u=x+\Delta_2 \: \Longrightarrow \: u\mid_{\alpha+\Delta_2=\Delta_3}^{\beta+\Delta_2=\Delta_4} \\&\therefore \: du=dx \\&=\displaystyle \dfrac{a}{c} \int_{\Delta_3}^{\Delta_4}
\dfrac{(u-\Delta_2)^n+ \Delta_1}{u} \: du \\&=\displaystyle \dfrac{a}{c} \int_{\Delta_3}^{\Delta_4}
\dfrac{(u-\Delta_2)^n}{u} \: du +
\dfrac{a\Delta_1}{c} \ln{( \dfrac{\Delta_4}{\Delta_3} )} \\&=\displaystyle \dfrac{a}{c} \int_{\Delta_3}^{\Delta_4}
\sum \limits_{k=0}^n \left[ \binom{n}{k}(u)^{n-k-1}(-1)^k(\Delta_2)^k \right] \: du +
\dfrac{a\Delta_1}{c} \ln{( \dfrac{\Delta_4}{\Delta_3} )} \\&=\displaystyle \dfrac{a}{c} \sum \limits_{k=0}^n \left( \left[ \binom{n}{k} (-1)^k(\Delta_2)^k \right]
\int_{\Delta_3}^{\Delta_4} (u)^{n-k-1} \: du \right) +
\dfrac{a\Delta_1}{c} \ln{( \dfrac{\Delta_4}{\Delta_3} )} \\&=\displaystyle \dfrac{a}{c} \sum \limits_{k=0}^n \left[ \binom{n}{k} \dfrac{(-1)^k(\Delta_2)^k}{n-k} \left[ (\Delta_4)^{n-k} - (\Delta_3)^{n-k} \right] \right] \: + \dfrac{a\Delta_1}{c} \ln{( \dfrac{\Delta_4}{\Delta_3} )} \\& \\&Note: \: \Delta_1=\dfrac{b}{a} \: , \: \Delta_2=\dfrac{d}{c} \\&\therefore \Delta_3=\alpha+\dfrac{d}{c}=\dfrac{\alpha c+d}{c} \\&\therefore \Delta_4=\beta+\dfrac{d}{c}=\dfrac{\beta c+d}{c} \\& \\&\therefore \dfrac{a\Delta_1}{c} \ln{( \dfrac{\Delta_4}{\Delta_3} )} \Longrightarrow \dfrac{b}{c} \ln{( \dfrac{\beta c+d}{\alpha c+d} )} \\&\therefore (\Delta_4)^{n-k} - (\Delta_3)^{n-k} \Longrightarrow
\dfrac{(\beta c+d)^{n-k}-(\alpha c+d)^{n-k}}{(c)^{n-k}} \\& \\&\end{align*}\begin{aligned}\therefore I=
\displaystyle a \sum \limits_{k=0}^n \left[ \binom{n}{k} \dfrac{(-d)^k[(\beta c+d)^{n-k}-(\alpha c+d)^{n-k}]}{(c)^{n-1}(n-k)} \right]
+ \dfrac{b}{c} \ln{( \dfrac{\beta c+d}{\alpha c+d} )}\end{aligned}

Inspired by BPRP's video on the Binomial Theorem:

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