一些不等积分的练习（持续更新）
最近会持续更新做一些这里面的题目，提供几个网站可以找答案：wolfram ，integral-calculator
1.求$\int \frac {1+x^2}{1+x^4}dx$
注意到以下事实$\int\frac{1}{x^2+a^2}dx=\frac{1}{a}arctan(\frac{x}{a})+C$
故$\int \frac {1+x^2}{1+x^4}dx\\=\int\frac{1}{2}(\frac{1}{x^2-\sqrt{2}x+1}+\frac{1}{x^2+\sqrt{2}x+1})dx\\=\frac{1}{2}(\int\frac{1}{(x-\frac{1}{\sqrt{2}})^2+\frac{1}{2}}d(x-\frac{1}{\sqrt2})+\int\frac{1}{(x+\frac{1}{\sqrt2})^2+\frac{1}{2}}d(x+\frac{1}{\sqrt2}))\\=\frac{\sqrt{2}}{2}(arctan(\sqrt2x-1)+arctan(\sqrt2x+1))+C$
2.求 $\int\frac{dx}{(x+1)(x^2+1)}$
注意到$\int \frac{dx}{a^2-x^2}=\frac{1}{a}arctanh(\frac{x}{a})+C=\frac{ln(|\frac{x+a}{-x+a}|)}{2a}+C\\arctanh(x)=\frac{1}{2}(ln(x+1)-ln(1-x))$
原式$=\int\frac{(x-1)dx}{x^4-1}\\=\int\frac{d(x^2)}{2(x^4-1)}-\int\frac{dx}{x^4-1}\\=-\frac{1}{2}arctanh(x^2)-\frac{1}{2}\int(\frac1{x^2-1}-\frac1{x^2+1})dx\\=-\frac{1}{2}arctanh(x^2)+\frac12arctanh(x)+\frac12arctan(x)+C$
3.求$\int \frac {dx}{(1+x^2)(1+x^3)}$
原式$=\int\frac{dx}{(1+x^2)(1+x)(1-x+x^2)}\\=-\frac{1}{3}\int\frac{2x-1}{x^2-x+1}dx+\frac{1}{2}\int \frac{x+1}{x^2+1}dx+\frac16\int\frac{1}{1+x}dx\\=-\frac{1}{3}ln|x^2-x+1|+\frac16ln|1+x|+\frac{1}{2}arctanx+\frac{1}{4}ln|x^2+1|+C$
4.求$\int \frac{dx}{\sqrt[3]{1-x^3}}$
令$t=\frac{x}{\sqrt[3]{1-x^3}}$，那么$dt=\frac{dx}{(\sqrt[3]{1-x^3})^4}$
原式$=\int (1-x^3)dt=\int \frac{dt}{t^3+1}=-\frac12ln|t^2-t+1|+\frac{1}{\sqrt3}arctan\frac{2t-1}{\sqrt3}+\frac{ln|t^3+1|}{3}+C$
5.求$\int \frac{dx}{\lambda+\sqrt{1-x^2}}$
当$|\lambda|>1$时,令$x=sint$
原式$=\int \frac{\lambda dx}{\lambda^2-1+x^2}-\int \frac{\sqrt{1-x^2}dx}{\lambda^2+x^2-1}=\frac{\lambda}{\sqrt{\lambda^2-1}}arctan{\frac{x}{\sqrt{\lambda^2-1}}}-\int\frac{cos^2t}{\lambda^2-cos^2t}dt=\frac{\lambda}{\sqrt{\lambda^2-1}}arctan{\frac{x}{\sqrt{\lambda^2-1}}}+t-\frac{arctan\frac{tant}{\sqrt{\lambda^2-1}}}{\sqrt{\lambda^2-1}}+C$
当$\lambda=1$时,显然$原式=arcsinx+C$
当$|\lambda|<1$时，原式$=\int \frac{\lambda dx}{\lambda^2-1+x^2}-\int \frac{\sqrt{1-x^2}dx}{\lambda^2+x^2-1}=\frac{\lambda}{\sqrt{1-\lambda^2}}arctanh\frac{x}{\sqrt{1-\lambda^2}}+t-\frac{arctanh\frac{tant}{\sqrt{1-\lambda^2}}}{\sqrt{1-\lambda^2}}+C$
6.求$\int \frac{dx}{asinx+bcosx}$
原式$=\frac{1}{\sqrt{a^2+b^2}}\int\frac{d(x+\phi)}{cos(x+\phi)}=\frac{1}{\sqrt{a^2+b^2}}ln|sec(x+\phi)+tan(x+\phi)|+C$
7.求$\int ln(\sqrt{x+1}+\sqrt{1-x})dx$
令$m=\sqrt{x+1},n=\sqrt{-x+1}$
原式$=\int ln(n+m)dx\\=xln(n+m)-\int xdln(n+m)$
记$I=\int xdln(n+m)$
注意到 $2x=m^2-n^2$ $m^2+n^2=2$
我们有$I$ $=\int \frac{m^2-n2}{2}*\frac{dn+dm}{m+n}\\=\int\frac{(dm+dn)(m-n)}{2}\\=\frac12\int (mdm-ndn)+\int \frac{ n^2d(\frac{n}{m})}{m^2+n^2}=\frac14(m^2-n^2)+arctan(\frac nm)+C$
将$n,m$带入，有$原式=xln(\sqrt{x+1}+\sqrt{x-1})-\frac{x}{2}-arctan(\frac{\sqrt{x-1}}{\sqrt{x+1}})+C$
8.求$\int ln(\sqrt{1-x}-\sqrt x)dx$ // unfinished
21.求$\int \sqrt{tanx+2}dx$
令$n=tanx,m=\sqrt{n^2+2}$
我们有 $mdm=ndn,\frac{dn}{n^2+1}=dx$
原式$=\int mdx\\=\int \frac{mdn}{n^2+1}\\=\int \frac{2mdn}{m^2+n^2}$
我们又注意到$\int \frac{mdn-ndm}{m^2+n^2}=arctan(\frac{n}{m})+C$
$\int \frac{mdn+ndm}{n^2+m^2}=\int\frac{mdn+\frac{n^2}{m}dn}{m^2+n^2}=\int\frac{dn}{m}$
不难发现$\frac{dn}{m}=\frac{dm}{n}=\frac{dn+dm}{m+n}$
所以 $\int \frac{mdn+ndm}{n^2+m^2}=\int\frac{mdn+\frac{n^2}{m}dn}{m^2+n^2}=\int\frac{dn}{m}=\int \frac{dm+dn}{n+m}=ln(n+m)+C$
故原式$=\int\frac{mdn-ndm}{m^2+n^2}+\int\frac{mdn+ndm}{n^2+m^2}=arctan(\frac{n}{m})+ln(n+m)+C=arctan(\frac{tanx}{\sqrt{tan^2x+2}})+ln(tanx+\sqrt{tan^2x+2})+C$
