文章目录

• 积分计算

积分计算

求导↔积分求导 \leftrightarrow 积分求导↔积分

tan2x+1=(sinxcosx)2+1=(sin2x+cos2xcos2x)=1cos2x=sec2xtan^2x+1=(\frac{sinx}{cosx})^2+1=(\frac{sin^2x+cos^2x}{cos^2x})=\frac{1}{cos^2x}=sec^2x \\~\\~tan2x+1=(cosxsinx​)2+1=(cos2xsin2x+cos2x​)=cos2x1​=sec2x  

不定积分公式【积分∫记得+C】\large 不定积分公式【积分\int记得+C】\\~不定积分公式【积分∫记得+C】 

1.∫secxdx=ln∣secx+tanx∣+C1.\large{\int secxdx=ln|secx+tanx|+C }\\~1.∫secxdx=ln∣secx+tanx∣+C 
2.∫cscxdx=ln∣cscx−cotx∣+C2. \large{\int cscxdx = ln|cscx-cotx|+C }\\~2.∫cscxdx=ln∣cscx−cotx∣+C 
3.∫1a2−x2dx=arcsinxa+C3.\large{ \int \frac{1}{\sqrt{a^2-x^2}}dx=arcsin\frac{x}{a}+C }\\~3.∫a2−x2​1​dx=arcsinax​+C 
4.∫1a2+x2dx=1aarctanxa+C4. \large{\int \frac{1}{a^2+x^2}dx = \frac{1}{a}arctan\frac{x}{a}+C}\\~4.∫a2+x21​dx=a1​arctanax​+C 
5.∫1x2±a2dx=ln∣x+x2±a2∣+C5. \large{\int \frac{1}{\sqrt{x^2 \pm a^2}}dx=ln|x+\sqrt{x^2 \pm a^2}| + C}\\~5.∫x2±a2​1​dx=ln∣x+x2±a2​∣+C 
6.∫11+exdx=x−ln(1+ex)+C6. \large{\int \frac{1}{1+e^x}dx=x-ln(1+e^x)+C} \\~6.∫1+ex1​dx=x−ln(1+ex)+C 
7.∫1a2−x2dx=12aln∣a+xa−x∣+C=−∫1x2−a2dx=−12aln∣x+ax−a∣+C7.\large{\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}ln|\frac{a+x}{a-x}|+C}=-\int \frac{1}{x^2-a^2}dx=-\frac{1}{2a}ln|\frac{x+a}{x-a}|+C \\~7.∫a2−x21​dx=2a1​ln∣a−xa+x​∣+C=−∫x2−a21​dx=−2a1​ln∣x−ax+a​∣+C 
8.∫x2+a2=x2x2+a2+a22ln(x+x2+a2)+C8. \large{\int \sqrt{x^2+a^2}= \frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2}ln(x+\sqrt{x^2+a^2})+C}\\~8.∫x2+a2​=2x​x2+a2​+2a2​ln(x+x2+a2​)+C 
9.∫x2−a2=x2x2−a2−a22ln(x+x2−a2)+C9. \large{\int \sqrt{x^2-a^2}= \frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2}ln(x+\sqrt{x^2-a^2})+C}\\~9.∫x2−a2​=2x​x2−a2​−2a2​ln(x+x2−a2​)+C 
10.∫a2−x2=x2a2−x2+a22arcsinxa10. \large{\int \sqrt{a^2-x^2}= \frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}arcsin\frac{x}{a}} \\~\\~\\~10.∫a2−x2​=2x​a2−x2​+2a2​arcsinax​   

因为有常数C,同一积分可有多种答案,可求导验证正确性:因为有常数C,同一积分可有多种答案,可求导验证正确性:因为有常数C,同一积分可有多种答案,可求导验证正确性:

1.(ln∣secx+tanx∣)′=1secx+tanx(secxtanx+sec2x)=secx(tanx+secx)secx+tanx=secx1.\large(ln|secx+tanx|)'=\frac{1}{secx+tanx}(secxtanx+sec^2x)=\frac{secx(tanx+secx)}{secx+tanx}=secx \\~1.(ln∣secx+tanx∣)′=secx+tanx1​(secxtanx+sec2x)=secx+tanxsecx(tanx+secx)​=secx 
2.(ln∣cscx−cotx∣)′=1cscx−cotx(−cscxcotx+csc2x)=cscx(−cotx+cscx)cscx−cotx=cscx2.\large(ln|cscx-cotx|)'=\frac{1}{cscx-cotx}(-cscxcotx+csc^2x)=\frac{cscx(-cotx+cscx)}{cscx-cotx}=cscx \\~2.(ln∣cscx−cotx∣)′=cscx−cotx1​(−cscxcotx+csc2x)=cscx−cotxcscx(−cotx+cscx)​=cscx 
3.(arcsinxa)′=11−(ax)2∗1a=1a2−x23.\large(arcsin\frac{x}{a})'=\frac{1}{\sqrt{1-(\frac{a}{x})^2}}*\frac{1}{a}=\frac{1}{\sqrt{a^2-x^2}} \\~3.(arcsinax​)′=1−(xa​)2​1​∗a1​=a2−x2​1​ 
4.(1aarctanxa)′=1a∗11+(xa)2=1a2+x24.\large(\frac{1}{a}arctan\frac{x}{a})'=\frac{1}{a}*\frac{1}{1+(\frac{x}{a})^2}=\frac{1}{a^2+x^2} \\~4.(a1​arctanax​)′=a1​∗1+(ax​)21​=a2+x21​ 
5.(ln∣x+x2±a2∣)′=1x+x2±a2∗(1+12x2±a2∗2x)=1x+x2±a2∗(x2±a2+xx2±a2)=1x2±a25.\large{(ln|x+\sqrt{x^2 \pm a^2}| )'= \frac{1}{x+\sqrt{x^2 \pm a^2}}*(1+\frac{1}{2\sqrt{x^2 \pm a^2}}*2x)=\frac{1}{x+\sqrt{x^2 \pm a^2}}*(\frac{\sqrt{x^2 \pm a^2 }+x}{\sqrt{x^2 \pm a^2}})=\frac{1}{\sqrt{x^2 \pm a^2}} }\\~5.(ln∣x+x2±a2​∣)′=x+x2±a2​1​∗(1+2x2±a2​1​∗2x)=x+x2±a2​1​∗(x2±a2​x2±a2​+x​)=x2±a2​1​ 

6.(x−ln(1+ex)+C)′=1−11+ex∗ex=1+ex−11+ex=11+ex6.\large{(x-ln(1+e^x)+C)'=1-\frac{1}{1+e^x}*e^x=\frac{1+e^x-1}{1+e^x}=\frac{1}{1+e^x} } \\~6.(x−ln(1+ex)+C)′=1−1+ex1​∗ex=1+ex1+ex−1​=1+ex1​ 
∫11+exdx多种积分解法:\large{\int \frac{1}{1+e^x}dx}多种积分解法:\\~∫1+ex1​dx多种积分解法: 
凑微分：∫11+exdx=∫1+ex−ex1+exdx=∫1−11+exd(1+ex)=x−ln(1+ex)+C凑微分：\large{\int \frac{1}{1+e^x}dx}=\large{\int \frac{1+e^x-e^x}{1+e^x}dx}=\large{\int 1-\frac{1}{1+e^x}d(1+e^x)}=x-ln(1+e^x)+C \\~凑微分：∫1+ex1​dx=∫1+ex1+ex−ex​dx=∫1−1+ex1​d(1+ex)=x−ln(1+ex)+C 
提公因子：∫11+exdx=∫1ex(e−x+1)dx=∫e−xe−x+1dx=−∫1e−x+1(e−x+1)=−ln(1+e−x)+C=−ln(ex+1ex)+C=−(ln(ex+1)−lnex)+C=x−ln(1+ex)+C提公因子：\large{\int \frac{1}{1+e^x}dx=\int \frac{1}{e^x(e^{-x}+1)}dx= \int \frac{e^{-x}}{e^{-x}+1}dx}=-\int \frac{1}{e^{-x}+1}(e^{-x}+1)=-ln(1+e^{-x})+C \\ =-ln(\frac{e^x+1}{e^{x}})+C=-(ln(e^x+1)-lne^x)+C=x-ln(1+e^x)+C \\~提公因子：∫1+ex1​dx=∫ex(e−x+1)1​dx=∫e−x+1e−x​dx=−∫e−x+11​(e−x+1)=−ln(1+e−x)+C=−ln(exex+1​)+C=−(ln(ex+1)−lnex)+C=x−ln(1+ex)+C 

换元法：∫11+exdx=∫exex(1+ex)dx=∫1ex(1+ex)dex令t=ex∫1t(1+t)dt=∫1t−11+tdt=lnex−ln(1+ex)+C=x−ln(1+ex)+C换元法：\large\int \frac{1}{1+e^x}dx=\int \frac{e^x}{e^x(1+e^x)}dx=\int \frac{1 }{e^x(1+e^x)}de^x ~~\frac{令t=e^x}{}~~ \int \frac{1}{t(1+t)}dt = \int \frac{1}{t}-\frac{1}{1+t}dt \\ =lne^x-ln(1+e^x)+C=x-ln(1+e^x)+C \\~换元法：∫1+ex1​dx=∫ex(1+ex)ex​dx=∫ex(1+ex)1​dex  令t=ex​  ∫t(1+t)1​dt=∫t1​−1+t1​dt=lnex−ln(1+ex)+C=x−ln(1+ex)+C