高数积分计算【合集】

积分计算

$求导 \rightleftarrows 积分$

$\large tan^2x+1=(\frac{sinx}{cosx})^2+1=(\frac{sin^2x+cos^2x}{cos^2x})=\frac{1}{cos^2x}=sec^2x \ \~\ 同理cot^2x+1=csc^2x\~$

$\int tan^2x dx = \int (sec^2x-1 )dx=tanx-x+c \~\~$

$\large 不定积分公式【积分\int记得+C】\~$ $3.和4.的常数a^2在前arc反三角函数,5.的x^2在前ln(组合); ~8.和9.为\sqrt{x^2}在前 \frac{x}{2} (本身)+ln(x+本身)，10.为\frac{x}{2} (本身)$ $\~\~$

$1.\large{\int secxdx=\int \frac{1}{cosx}dx=\int \frac{dx}{cosx}=ln|secx+tanx|+C }\~$ $2. \large{\int cscxdx =\int \frac{1}{sinx}dx =\int \frac{dx}{sinx}= ln|cscx-cotx|+C }\~$ $3.\large{ \int \frac{1}{\sqrt{a^2-x^2}}dx=arcsin\frac{x}{a}+C }\~$ $4. \large{\int \frac{1}{a^2+x^2}dx = \frac{1}{a}arctan\frac{x}{a}+C}\~$ $5. \large{\int \frac{1}{\sqrt{x^2 \pm a^2}}dx=ln|x+\sqrt{x^2 \pm a^2}| + C}\~$ $6. \large{\int \frac{1}{1+e^x}dx=x-ln(1+e^x)+C=-ln(1+e^{-x})+C }\~$ $7.\large{\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}ln|\frac{a+x}{a-x}|+C=-[\int \frac{1}{x^2-a^2}dx]=-[\frac{1}{2a}ln|\frac{x-a}{x+a}|]+C }\~$ $8. \large{\int \sqrt{x^2+a^2}= \frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2}ln(x+\sqrt{x^2+a^2})+C}\~$ $9. \large{\int \sqrt{x^2-a^2}= \frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2}ln(x+\sqrt{x^2-a^2})+C}\~$ $10. \large{\int \sqrt{a^2-x^2}= \frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}arcsin\frac{x}{a}+C} \~\~\~$

$因为有常数C,同一积分可有多种答案,可求导验证正确性:$

$1.\large(ln|secx+tanx|)'=\frac{1}{secx+tanx}(secxtanx+sec^2x)=\frac{secx(tanx+secx)}{secx+tanx}=secx \~$ $2.\large(ln|cscx-cotx|)'=\frac{1}{cscx-cotx}(-cscxcotx+csc^2x)=\frac{cscx(-cotx+cscx)}{cscx-cotx}=cscx \~$ $3.\large(arcsin\frac{x}{a})'=\frac{1}{\sqrt{1-(\frac{a}{x})^2}}\frac{1}{a}=\frac{1}{\sqrt{a^2-x^2}} \~$ $4.\large(\frac{1}{a}arctan\frac{x}{a})'=\frac{1}{a}\frac{1}{1+(\frac{x}{a})^2}=\frac{1}{a^2+x^2} \~$ $5.\large{(ln|x+\sqrt{x^2 \pm a^2}| )'= \frac{1}{x+\sqrt{x^2 \pm a^2}}*(1+\frac{1}{2\sqrt{x^2 \pm a^2}}2x)=\frac{1}{x+\sqrt{x^2 \pm a^2}}(\frac{\sqrt{x^2 \pm a^2 }+x}{\sqrt{x^2 \pm a^2}})=\frac{1}{\sqrt{x^2 \pm a^2}} }\~$

$6.\large{(x-ln(1+e^x)+C)'=1-\frac{1}{1+e^x}*e^x=\frac{1+e^x-1}{1+e^x}=\frac{1}{1+e^x} } \~$ $\large{\int \frac{1}{1+e^x}dx}多种积分解法:\~$ $凑微分：\large{\int \frac{1}{1+e^x}dx}=\large{\int \frac{1+e^x-e^x}{1+e^x}dx}=\large{\int 1dx-\int \frac{1}{1+e^x}d(1+e^x)}=x-ln(1+e^x)+C \~$ $提公因子：\large{\int \frac{1}{1+e^x}dx=\int \frac{1}{e^x(e^{-x}+1)}dx= \int \frac{e^{-x}}{e^{-x}+1}dx}=-\int \frac{1}{e^{-x}+1}(e^{-x}+1)=-ln(1+e^{-x})+C \ =-ln(\frac{e^x+1}{e^{x}})+C=-(ln(e^x+1)-lne^x)+C=x-ln(1+e^x)+C \~$

$换元法：\large\int \frac{1}{1+e^x}dx=\int \frac{e^x}{e^x(1+e^x)}dx=\int \frac{1 }{e^x(1+e^x)}de^x \frac{令t=e^x}{} \int \frac{1}{t(1+t)}dt = \int \frac{1}{t}-\frac{1}{1+t}dt \ =lne^x-ln(1+e^x)+C=x-ln(1+e^x)+C \~$

$7.\large(\frac{1}{2a}ln|\frac{a+x}{a-x}|)'=\frac{1}{2a}(ln|a+x|-ln|a-x|)'=\frac{1}{2a}(\frac{1}{a+x}1 -\frac{1}{a-x}(-1)) [\small{复合函数求导}] \ \large=\frac{1}{2a}(\frac{(a-x)+(a+x)}{a^2-x^2})=\frac{1}{2a}(\frac{2a}{a^2-x^2})=\frac{1}{a^2-x^2} \~$

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

$8.\large (\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2}ln(x+\sqrt{x^2+a^2}~)~)' \ =\frac{1}{2} \sqrt{x^2+a^2}+\frac{x}{2} (\frac{1}{2}{(x^2+a^2)}^{-\frac{1}{2}}2x)+\frac{a^2}{2} \frac{1}{x+\sqrt{x^2+a^2}}(1+\frac{1}{2}{(x^2+a^2)}^{-\frac{1}{2}}*2x) \ =\frac{1}{2} \sqrt{x^2+a^2}+\frac{x^2}{2 \sqrt{x^2+a^2}}+\frac{a^2}{2}(\frac{1}{x+\sqrt{x^2+a^2}})(\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}})=\frac{1}{2}\sqrt{x^2+a^2}+\frac{x^2}{2 \sqrt{x^2+a^2}}+\frac{a^2}{2 \sqrt{x^2+a^2}} \ =\frac{1}{2}\sqrt{x^2+a^2}+\frac{x^2+a^2}{2 \sqrt{x^2+a^2}}=\frac{1}{2}\sqrt{x^2+a^2}+\frac{1}{2}\sqrt{x^2+a^2}=\sqrt{x^2+a^2} \~$

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

$10.\large(\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}arcsin\frac{x}{a})'=\frac{1}{2} \sqrt{a^2-x^2}+\frac{x}{2} ( \frac{1}{2\sqrt{a^2-x^2}}(-2x))+\frac{a^2}{2}(\frac{1}{\sqrt{1-(\frac{x}{a})^2}}(\frac{1}{a})) \ =\frac{1}{2} \sqrt{a^2-x^2}+\frac{-x^2}{2\sqrt{a^2-x^2}}+\frac{a^2}{2\sqrt{a^2-x^2}} =\frac{1}{2} \sqrt{a^2-x^2}+ \frac{a^2-x^2}{2\sqrt{a^2-x^2}} \ =\frac{1}{2} \sqrt{a^2-x^2}+\frac{1}{2} \sqrt{a^2-x^2}= \sqrt{a^2-x^2} \~\~\~$

$\large 换元法【两类】$ $第一类换元法：凑微分(复合函数逆过程)$

$如(f(g(x)))'=f'(g(x))*g'(x)$

$对应积分\int f'(g(x))*g'(x)dx=\int f'(g(x))dg(x)$ $简单换元看出：\frac{令t=g(x)}{} \int f'(t)dt=f(t)+c$

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

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

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

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

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

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

$积累变型：\ \large \int \frac{dx}{a^2sin^2x+b^2cos^2x}=\int \frac{dx}{cos^2x(a^2tan^2x+b^2)}=\int \frac{sec^2x}{(a^2tan^2x+b^2)}dx=\int \frac{1}{(a^2tan^2x+b^2)}d(tanx) \ =\frac{1}{a}\int \frac{1}{(a^2tan^2x+b^2)}d(atanx)=\frac{1}{a}(\frac{1}{b}arctanx\frac{atanx}{b})+C ~~~[a,b\ne 0]\~$

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

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

$\~\~\~\~$ $第二类换元法$

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

$三角代换推导公式:$

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

$\~$ $反三角函数：$ $\int \frac{dx}{x\sqrt{x^2-1}}~\frac{令x=sect}{}~\int \frac{secttant}{secttant}dt~~[(sect)'=secttant;dx=secttantdt] \ =\int 1dt=t+c ~~[x=sect=\frac{1}{cost} \to~t=arccos\frac{1}{x}] \ =arccos\frac{1}{x}+c \~$

$分部积分法：$ $(uv)'=u'v+uv' (逆运算推导分部积分)：$ $\ \int(uv)'dx=\int u'vdx+\int uv'dx \ uv = \int vdu+\int udv \ \int udv =uv-\int vdu \~$ $即：\int udv=uv-\int vdu$ $\~$

$[表格法：对角相连，正负相间] +(上导下积)\~$ $\int xsinxdx$ $求导:x ~\to~ 1 ~\to~ 0$ $\searrow +~\searrow -$ $积分:sinx \to -cosx \to -sinx$ $\int xsinxdx=-x cosx+sinx+c \~$ $\int xe^xdx$ $求导:~x \to~ 1 ~\to~ 0$ $\searrow +\searrow -$ $积分:e^x \to e^x\to~~e^x$ $\int xe^xdx=xe^x-e^x$

$\~$ $记住~或~能推导 : \ \int sec^3xdx,\int csc^3xdx,\int e^{ax}sin(bx)dx,\int e^{ax}cos(bx)dx,\int sin(lnx)dx,\int cos(lnx)dx~~[令t=lnx] \~\ \large\int e^{ax}sin(bx)dx= \frac{1}{a^2+b^2} \begin{vmatrix} (e^{ax})' & (sin(bx))' \ e^{ax} & sin(bx) \ \end{vmatrix} +C ~~~~~~\small[化成行列式] \~\ \large= \frac{1}{a^2+b^2}(ae^{ax}*sinbx-e^{ax}*bcosbx)+C \ \~\~$

$\large 有理函数积分$

$例1~ 求 \int \frac{x}{(x+1)(x+2)(x+3)}dx$

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

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

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

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

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

$\large 三角函数积分 \~$ $\int sin^2xdx、\int cos^2xdx偶数次幂，降幂$ $\int sin^2xdx=\frac{1}{2}\int \frac{1-cos2x}{2}d2x=\frac{1}{2}x-\frac{1}{4}\sin2x+c [求导检查正确性]\~$

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

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

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

$\large \red{不定积分+C,定积分没有常数C}$