Step By Step Calculus » 15.6 - Use of Integral Forms

show_hiddenExpand All [+]
Synopsis
We build upon the previous sections by combining some of the forms studied there with a change of variables, creating more general forms. We also extend these ideas by including hyperbolic substitutions. It is useful to introduce some common substitution forms for recognizing seemingly complicated antiderivatives:
$\displaystyle \newcommand{\T}{\rule{0pt}{2.8ex}} \newcommand{\B}{\rule[-1.9ex]{0pt}{0pt}} \begin{array}{l|l|l|l|l} \B&\text{Form of }f & \text{Use} & \text{When } &\int f(g(x))g^{\prime }(x)dx \\ \hline \T\B\text{\bf A. }&ag^{r} & & r \neq -1 & \frac{a}{r+1}g^{r+1}(x)+c \\ \T\B\text{\bf B. }&\frac{a}{g-b} & \ln & g(x)-b \neq 0 & a\ln \left| g(x)-b\right| +c \\ \T\B\text{\bf C. }&a^{g} & & a>0 & \frac{a^{g(x)}}{\ln (a)}+c \\ \T\B\text{\bf D. }&\frac{g}{g^2+b} & \ln & b\in \mathbb{R}, g^2+b \neq 0 & \frac{1}{2}\ln \left| g^2+b \right|+c\\ \T\B\text{\bf E. }&\frac{a}{b+g^{2}} & \tan ^{-1}\text{ or cot}^{-1} & b>0 & \frac{a}{% \sqrt{b}}\tan ^{-1}\left( \frac{g(x)}{\sqrt{b}}\right) +c \\ \T\B\text{\bf F. }&\frac{a}{\sqrt{b-g^{2}}} & \sin ^{-1}\text{ or cos}^{-1} & b>0 & a\sin ^{-1}\left( \frac{g(x)}{\sqrt{b}}\right) +c \\ \T\B\text{\bf G. }&\frac{a}{\sqrt{g^2-b}} & \cosh^{-1} & b>0 & a\cosh^{-1}\left( \frac{g(x)}{\sqrt{b}}\right) +c \\ \T\B\text{\bf H. }&\frac{a}{b-g^{2}} & \tanh ^{-1}\text{ or coth}^{-1} & b>0 & \frac{a}{% \sqrt{b}}\tanh ^{-1}\left( \frac{g(x)}{\sqrt{b}}\right) +c \\ \T\B\text{\bf I. }&\frac{a}{\sqrt{b+g^{2}}} & \sinh {}^{-1} & b>0 & a\sinh ^{-1}\left( \frac{g(x)}{\sqrt{b}}\right) +c\\ \T\B\text{\bf J. }&2g^{2n+1}\left( g^2+a\right)^{\frac{m}{2}} & & n\in\mathbb{N},m\in\mathbb{Z},a\in\mathbb{R}& \sum_{k=0}^n {n\choose k} \frac{\left( g^2+a\right)^{k+1+\frac{m}{2}}}{k+1+\frac{m}{2}}(-a)^{n-k}+c\\ \T\B\text{\bf K. }&\frac{(3-2r)g^2+d^2}{\left( g^2+d^2\right)^r} & & r=2,3,4,\cdots & \frac{g}{\left( g^2+d^2\right)^{r-1}} +c \\\hline \end{array} . $