积分计算

$求导\rightleftarrows 积分$

求导	积分
$(t{)}^{\prime }=1$	$\int tdt=\frac{1}{2}{t}^2+c$
$(\frac{1}{x}{)}^{\prime }=-\frac{1}{x^2}$	$\int \frac{1}{x^2}dx=-\frac{1}{x}+c$
$(lnx{)}^{\prime }=\frac{1}{x}$	$\int \frac{1}{x}dx=\int \frac{dx}{x}=ln|x|+c$
$(\sqrt{x}{)}^{\prime }=\frac{1}{2\sqrt{x}}$	$\int \frac{1}{\sqrt{x}}dx=2\sqrt{x}+c$
$(a^x{)}^{\prime }=a^{x}lna$	$\int a^{x}dx=\frac{a^x}{lna}+c$
$(sinx{)}^{\prime }=cosx$	$\int cosxdx=sinx+c$
$(tanx{)}^{\prime }=se{c}^{2}x$	$\int se{c}^{2}xdx=tanx+c$
$(secx{)}^{\prime }=(\frac{1}{cosx}{)}^{\prime }=secxtanx=-\frac{sinx}{co{s}^{2}x}$	$\int secxtanxdx=secx+c$
$(cosx{)}^{\prime }=-sinx$	$\int sinxdx=-cosx+c$
$(cotx{)}^{\prime }=-cs{c}^{2}x$	$\int cs{c}^{2}xdx=-cotx+c$
$(cscx{)}^{\prime }=(\frac{1}{sinx}{)}^{\prime }=-csccotx=-\frac{cosx}{si{n}^{2}x}$	$\int csccotdx=-cscx+c$
$(arctanx{)}^{\prime }=\frac{1}{1+x^2}$	$\int \frac{1}{1+x^2}dx=arctanx+c$
$(arccotx{)}^{\prime }=-\frac{1}{1+x^2}$	$\int -\frac{1}{1+x^2}dx=arccotx+c$
$(arcsinx{)}^{\prime }=\frac{1}{\sqrt{1-x^2}}$	$\int \frac{1}{\sqrt{1-x^2}}dx=arcsinx+c$
$(arccosx{)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}$	$\int -\frac{1}{\sqrt{1-x^2}}dx=arccosx+c$

$$\begin{gathered} ta{n}^{2}x+1=(\frac{sinx}{cosx}{)}^2+1=(\frac{si{n}^{2}x+co{s}^{2}x}{co{s}^{2}x})=\frac{1}{co{s}^{2}x}=se{c}^{2}x \\ 同理co{t}^{2}x+1=cs{c}^{2}x \end{gathered}$$

$\int ta{n}^{2}xdx=\int (se{c}^{2}x-1)dx=tanx-x+c$

$不定积分公式【积分\int 记得+C】$
$3.和4.的常数a^2在前arc反三角函数,5.的x^2在前ln(组合);8.和9.为\sqrt{x^2}在前\frac{x}{2}(本身)+ln(x+本身)，10.为\frac{x}{2}(本身)$
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$1.\int secxdx=\int \frac{1}{cosx}dx=\int \frac{dx}{cosx}=ln|secx+tanx|+C$
$2.\int cscxdx=\int \frac{1}{sinx}dx=\int \frac{dx}{sinx}=ln|cscx-cotx|+C$
$3.\int \frac{1}{\sqrt{a^2-x^2}}dx=arcsin\frac{x}{a}+C$
$4.\int \frac{1}{a^2+x^2}dx=\frac{1}{a}arctan\frac{x}{a}+C$
$5.\int \frac{1}{\sqrt{x^2\pm a^2}}dx=ln|x+\sqrt{x^2\pm a^2}|+C$
$6.\int \frac{1}{1+e^x}dx=x-ln(1+e^x)+C=-ln(1+e^{-x})+C$
$7.\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}ln|\frac{a+x}{a-x}|+C=-[\int \frac{1}{x^2-a^2}dx]=-[\frac{1}{2a}ln|\frac{x-a}{x+a}|]+C$
$8.\int \sqrt{x^2+a^2}=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}ln(x+\sqrt{x^2+a^2})+C$
$9.\int \sqrt{x^2-a^2}=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}ln(x+\sqrt{x^2-a^2})+C$
$10.\int \sqrt{a^2-x^2}=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}arcsin\frac{x}{a}+C$

$因为有常数C,同一积分可有多种答案,可求导验证正确性:$

$1.(ln|secx+tanx|{)}^{\prime }=\frac{1}{secx+tanx}(secxtanx+se{c}^{2}x)=\frac{secx(tanx+secx)}{secx+tanx}=secx$
$2.(ln|cscx-cotx|{)}^{\prime }=\frac{1}{cscx-cotx}(-cscxcotx+cs{c}^{2}x)=\frac{cscx(-cotx+cscx)}{cscx-cotx}=cscx$
$3.(arcsin\frac{x}{a}{)}^{\prime }=\frac{1}{\sqrt{1-(\frac{a}{x}{)}^2}}\ast \frac{1}{a}=\frac{1}{\sqrt{a^2-x^2}}$
$4.(\frac{1}{a}arctan\frac{x}{a}{)}^{\prime }=\frac{1}{a}\ast \frac{1}{1+(\frac{x}{a}{)}^2}=\frac{1}{a^2+x^2}$
$5.(ln|x+\sqrt{x^2\pm a^2}|{)}^{\prime }=\frac{1}{x+\sqrt{x^2\pm a^2}}\ast (1+\frac{1}{2\sqrt{x^2\pm a^2}}\ast 2x)=\frac{1}{x+\sqrt{x^2\pm a^2}}\ast (\frac{\sqrt{x^2\pm a^2}+x}{\sqrt{x^2\pm a^2}})=\frac{1}{\sqrt{x^2\pm a^2}}$

$6.(x-ln(1+e^x)+C{)}^{\prime }=1-\frac{1}{1+e^x}\ast e^x=\frac{1+e^x-1}{1+e^x}=\frac{1}{1+e^x}$
$\int \frac{1}{1+e^x}dx多种积分解法:$
$凑微分：\int \frac{1}{1+e^x}dx=\int \frac{1+e^x-e^x}{1+e^x}dx=\int 1dx-\int \frac{1}{1+e^x}d(1+e^x)=x-ln(1+e^x)+C$
$$\begin{gathered} 提公因子：\int \frac{1}{1+e^x}dx=\int \frac{1}{e^x(e^{-x}+1)}dx=\int \frac{e^{-x}}{e^{-x}+1}dx=-\int \frac{1}{e^{-x}+1}(e^{-x}+1)=-ln(1+e^{-x})+C \\ =-ln(\frac{e^x+1}{e^x})+C=-(ln(e^x+1)-ln{e}^x)+C=x-ln(1+e^x)+C \end{gathered}$$

$$\begin{gathered} 换元法：\int \frac{1}{1+e^x}dx=\int \frac{e^x}{e^x(1+e^x)}dx=\int \frac{1}{e^x(1+e^x)}d{e}^x\frac{令t=e^x}{}\int \frac{1}{t(1+t)}dt=\int \frac{1}{t}-\frac{1}{1+t}dt \\ =ln{e}^x-ln(1+e^x)+C=x-ln(1+e^x)+C \end{gathered}$$

$$\begin{gathered} 7.(\frac{1}{2a}ln|\frac{a+x}{a-x}|{)}^{\prime }=\frac{1}{2a}(ln|a+x|-ln|a-x|{)}^{\prime }=\frac{1}{2a}(\frac{1}{a+x}\ast 1-\frac{1}{a-x}\ast (-1))[复合函数求导] \\ =\frac{1}{2a}(\frac{(a-x)+(a+x)}{a^2-x^2})=\frac{1}{2a}(\frac{2a}{a^2-x^2})=\frac{1}{a^2-x^2} \end{gathered}$$

$\int \frac{1}{x^2-a^2}dx=\int \frac{1}{(x+a)(x-a)}dx=\int \frac{1}{2a}[\frac{1}{x-a}-\frac{1}{x+a}]dx=\frac{1}{2a}ln|\frac{a+x}{a-x}|$

$$\begin{gathered} 8.(\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}ln(x+\sqrt{x^2+a^2}){)}^{\prime } \\ =\frac{1}{2}\sqrt{x^2+a^2}+\frac{x}{2}(\frac{1}{2}{(x^2+a^2)}^{-\frac{1}{2}}\ast 2x)+\frac{a^2}{2}\frac{1}{x+\sqrt{x^2+a^2}}\ast (1+\frac{1}{2}{(x^2+a^2)}^{-\frac{1}{2}}\ast 2x) \\ =\frac{1}{2}\sqrt{x^2+a^2}+\frac{x^2}{2\sqrt{x^2+a^2}}+\frac{a^2}{2}(\frac{1}{x+\sqrt{x^2+a^2}})(\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}})=\frac{1}{2}\sqrt{x^2+a^2}+\frac{x^2}{2\sqrt{x^2+a^2}}+\frac{a^2}{2\sqrt{x^2+a^2}} \\ =\frac{1}{2}\sqrt{x^2+a^2}+\frac{x^2+a^2}{2\sqrt{x^2+a^2}}=\frac{1}{2}\sqrt{x^2+a^2}+\frac{1}{2}\sqrt{x^2+a^2}=\sqrt{x^2+a^2} \end{gathered}$$

$8.(\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}ln(x+\sqrt{x^2-a^2}){)}^{\prime }=\sqrt{x^2-a^2}$

$$\begin{gathered} 10.(\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}arcsin\frac{x}{a}{)}^{\prime }=\frac{1}{2}\sqrt{a^2-x^2}+\frac{x}{2}(\frac{1}{2\sqrt{a^2-x^2}}\ast (-2x))+\frac{a^2}{2}(\frac{1}{\sqrt{1-(\frac{x}{a}{)}^2}}\ast (\frac{1}{a})) \\ =\frac{1}{2}\sqrt{a^2-x^2}+\frac{-x^2}{2\sqrt{a^2-x^2}}+\frac{a^2}{2\sqrt{a^2-x^2}}=\frac{1}{2}\sqrt{a^2-x^2}+\frac{a^2-x^2}{2\sqrt{a^2-x^2}} \\ =\frac{1}{2}\sqrt{a^2-x^2}+\frac{1}{2}\sqrt{a^2-x^2}=\sqrt{a^2-x^2} \end{gathered}$$

$换元法【两类】$
$第一类换元法：凑微分(复合函数逆过程)$

$如(f(g(x)){)}^{\prime }=f^{\prime }(g(x))\ast g^{\prime }(x)$

$对应积分\int f^{\prime }(g(x))\ast g^{\prime }(x)dx=\int f^{\prime }(g(x))dg(x)$
$简单换元看出：\frac{令t=g(x)}{}\int f^{\prime }(t)dt=f(t)+c$

$又如：出题\int e^{\Delta }{\Delta }^{\prime }dx,做题转换成\int e^{\Delta }d\Delta$

$\int e^{x^2}xdx=\frac{1}{2}\int e^{x^2}\ast 2xdx=\frac{1}{2}\int e^{x^2}d{x}^2=\frac{1}{2}{e}^{x^2}+c$
$\int e^{\sqrt{x}}\frac{1}{\sqrt{x}}dx=2\int e^{\sqrt{x}}\frac{1}{2\sqrt{x}}dx=2\int e^{\sqrt{x}}d\sqrt{x}=2e^{\sqrt{x}}+c$

$\int e^{arctanx}\frac{1}{1+x^2}dx=\int e^{arctanx}d(arctanx)=e^{arctanx}+c$

$补：(lnlnx{)}^{\prime }=\frac{1}{lnx}\frac{1}{x}=\frac{1}{xlnx}$

$\int \frac{1}{xlnxlnlnx}dx=\frac{\frac{1}{xlnx}}{lnlnx}dx=\frac{1}{lnlnx}d(lnlnx)=ln(lnlnx)+c$

$\int \frac{mx+n}{a{x}^2+bx+c}先分母求导(a{x}^2+bx+c{)}^{\prime }=2ax+b,则分子凑出2ax+b$

$$\begin{gathered} 积累变型： \\ \int \frac{dx}{a^{2}si{n}^{2}x+b^{2}co{s}^{2}x}=\int \frac{dx}{co{s}^{2}x(a^{2}ta{n}^{2}x+b^2)}=\int \frac{se{c}^{2}x}{(a^{2}ta{n}^{2}x+b^2)}dx=\int \frac{1}{(a^{2}ta{n}^{2}x+b^2)}d(tanx) \\ =\frac{1}{a}\int \frac{1}{(a^{2}ta{n}^{2}x+b^2)}d(atanx)=\frac{1}{a}(\frac{1}{b}arctanx\frac{atanx}{b})+C[a,b\ne 0] \end{gathered}$$

$\int \frac{1-lnx}{(x-lnx{)}^2}dx=\int \frac{\frac{1-lnx}{x^2}}{(1-\frac{lnx}{x}{)}^2}dx=-\int \frac{1}{(1-\frac{lnx}{x}{)}^2}d(1-\frac{lnx}{x})=(1-\frac{lnx}{x}{)}^{-1}+C$

$凑微分复杂看不出时，试试某部分求导[出题逆向]$
$$\begin{gathered} \int \frac{e^{tan\frac{1}{x}}se{c}^2\frac{1}{x}}{x^2}dx \\ 尝试(tan\frac{1}{x}{)}^{\prime }=se{c}^2\frac{1}{x}\ast (-\frac{1}{x^2})=-\frac{se{c}^2\frac{1}{x}}{x^2} \end{gathered}$$
$原式=\int e^{tan\frac{1}{x}}dtan\frac{1}{x}=e^{tan\frac{1}{x}}+c$

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$第二类换元法$

$三角代换：【注意换回x】$
$\sqrt{a^2-x^2},令x=a\sin t或a\cos t$
$\sqrt{a^2+x^2},令x=a\tan t或a\cot t$
$\sqrt{x^2-a^2},令x=a\sec t或a\csc t$
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$三角代换推导公式:$

$根式代换：令t=\sqrt{根式}$
$$\begin{gathered} \int \frac{\sqrt{x^3}}{\sqrt{x}+1}dx\frac{令t=\sqrt{x}}{}\int \frac{t^3}{t+1}2tdt[t^2=x,2tdt=dx] \\ =2\int \frac{t^4}{t+1}dt=2\int [(t-1)(t^2+1)+\frac{1}{t+1}]dt-\int \frac{t}{t+1}dt \\ 2\int \frac{t^4-t+t^3-t^3+t^2-t^2+t-t+1-1}{t+1}dt(多项式除法，凑分母倍数) \\ =2\int \frac{t^3(t+1)-t^2(t+1)+t(t+1)-2(t+1)+2}{t+1}dt \\ =2\int t^3-t^2+t-2+\frac{2}{t+1}dt=\frac{1}{2}{t}^4-\frac{2}{3}{t}^3+t^2-2t+2ln|t+1|+c \\ =\frac{1}{2}{x}^2-\frac{2}{3}\sqrt{x^3}+x-4\sqrt{x}+4ln(\sqrt{x}+1)+c[lnx,x>0] \end{gathered}$$

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$反三角函数：$
$$\begin{gathered} \int \frac{dx}{x\sqrt{x^2-1}}\frac{令x=sect}{}\int \frac{sect\ast tant}{sect\ast tant}dt[(sect{)}^{\prime }=sect\ast tant;dx=sect\ast tantdt] \\ =\int 1dt=t+c[x=sect=\frac{1}{cost}\to t=arccos\frac{1}{x}] \\ =arccos\frac{1}{x}+c \end{gathered}$$

$分部积分法：$
$(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }(逆运算推导分部积分)：$
$$\begin{gathered} \int (uv{)}^{\prime }dx=\int u^{\prime }vdx+\int u{v}^{\prime }dx \\ uv=\int vdu+\int udv \\ \int udv=uv-\int vdu \end{gathered}$$
$即：\int udv=uv-\int vdu$
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$[表格法：对角相连，正负相间]+(上导下积)$
$\int xsinxdx$
$求导:x\to 1\to 0$
$\searrow +\searrow -$
$积分:sinx\to -cosx\to -sinx$
$\int xsinxdx=-xcosx+sinx+c$
$\int x{e}^{x}dx$
$求导:x\to 1\to 0$
$\searrow +\searrow -$
$积分:e^x\to e^x\to e^x$
$\int x{e}^{x}dx=x{e}^x-e^x$

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$$\begin{gathered} 记住或能推导: \\ \int se{c}^{3}xdx,\int cs{c}^{3}xdx,\int e^{ax}sin(bx)dx,\int e^{ax}cos(bx)dx,\int sin(lnx)dx,\int cos(lnx)dx[令t=lnx] \\ \int e^{ax}sin(bx)dx=\frac{1}{a^2+b^2}\left|\begin{array}{cc}(e^{ax}{)}^{\prime }& (sin(bx){)}^{\prime }\\ e^{ax}& sin(bx)\end{array}\right|+C[化成行列式] \\ =\frac{1}{a^2+b^2}(a{e}^{ax}\ast sinbx-e^{ax}\ast bcosbx)+C \end{gathered}$$

$有理函数积分$

$例1求\int \frac{x}{(x+1)(x+2)(x+3)}dx$

$设原式=\int (\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3})dx=\int \frac{x}{(x+1)(x+2)(x+3)}dx$

$由待定系数法：A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)=x$
$$\begin{gathered} [特殊值法] \\ 令x=-1,2A=-1,即A=-\frac{1}{2}; \\ 令x=-2,-B=-2,即B=2; \\ 令x=-3,2C=-3,即C=-\frac{3}{2} \end{gathered}$$

$原式=\int (\frac{-\frac{1}{2}}{x+1}+\frac{2}{x+2}+\frac{-\frac{3}{2}}{x+3})dx=-\frac{1}{2}ln|x+1|+2ln|x+2|-\frac{3}{2}ln|x+3|+c$

$例2:分母括号内能分解尽量分解，如$
$$\begin{gathered} \int \frac{1}{(x+1)(x^2-1)}dx=\int \frac{1}{(x+1{)}^2(x-1)}dx \\ =\int [\frac{A}{x+1}+\frac{B}{(x+1{)}^2}+\frac{C}{x-1}]dx==\int \frac{A(x+1)(x-1)+B(x-1)+C(x+1{)}^2}{(x+1{)}^2(x-1)}dx \\ 待定系数法：A(x+1)(x-1)+B(x-1)+C(x+1{)}^2=1[特殊值法] \\ 令x=-1,-2B=1,B=-\frac{1}{2}; \\ 令x=-1,4C=1,C=\frac{1}{4}; \\ 令x=0,-A-B+C=1,A=-B+C-1=-\frac{1}{4}原式=-\frac{1}{4}ln|x+1|+\frac{1}{2}\frac{1}{x+1}+\frac{1}{4}ln|x-1|+C \end{gathered}$$

$$\begin{gathered} 最简时分母项中幂大于1，依据有理函数分解原则(能互相抵消成分子必须具备的可能) \\ \int \frac{1}{(x+1)(x^2+1)}dx=\int \frac{A}{x+1}+\frac{Bx+C}{x^2+1}dx[\leftarrow 此题重点处] \\ 待定系数法(通分)： \\ A(x^2+1)+(Bx+C)(x+1)=1 \\ A{x}^2+A+B{x}^2+Bx+Cx+C=1 \\ (A+B)x^2+(B+C)x+A+C=1 \\ 则A+B=0,B+C=0,A+C=1即C=\frac{1}{2},B=-\frac{1}{2}，A=\frac{1}{2} \\ 原式=\int \frac{\frac{1}{2}}{x+1}+\frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1}dx=\frac{1}{2}\int \frac{1}{x+1}+\frac{-x+1}{x^2+1}dx \\ =\frac{1}{2}\int \frac{1}{x+1}+\frac{1}{x^2+1}dx-\frac{1}{2}\ast \frac{1}{2}\int \frac{1}{x^2+1}d(x^2+1) \\ =\frac{1}{2}ln|1+x|+\frac{1}{2}arctanx+-\frac{1}{4}ln(x^2+1)+c \end{gathered}$$

$三角函数积分$
$\int si{n}^{2}xdx、\int co{s}^{2}xdx偶数次幂，降幂$
$\int si{n}^{2}xdx=\frac{1}{2}\int \frac{1-cos2x}{2}d2x=\frac{1}{2}x-\frac{1}{4}\sin 2x+c[求导检查正确性]$

$\int si{n}^{3}xdx、\int co{s}^{3}xdx奇数次幂凑微分$
$\int si{n}^{3}xdx=-\int si{n}^{2}xdcosx=-\int (1-co{s}^{2}x)dcosx=\int (co{s}^{2}x-1)dcosx=\frac{1}{3}co{s}^{3}x-cosx+c$

$万能替换公式：$
$令u=tan\frac{x}{2}\to x=2arctanu,dx=\frac{2du}{1+u^2}$
$$\begin{gathered} sinx=\frac{2tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}=\frac{2u}{1+u^2} \\ cosx=\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}=\frac{1-u^2}{1+u^2} \end{gathered}$$

$\int \frac{1}{1+sinx+cosx}dx=\int \frac{\frac{2du}{1+u^2}}{1+\frac{2u}{1+u^2}+\frac{1-u^2}{1+u^2}}=\frac{2}{2+2u}du=\frac{1}{1+u}du=ln|1+u|+c=ln|1+tan\frac{x}{2}|+c$

递推+重点补充

$不定积分+C,定积分没有常数C$