100道积分公式证明（71-100）
100道积分公式证明（71-100）
最新推荐文章于 2025-01-26 14:30:34 发布
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最新推荐文章于 2025-01-26 14:30:34 发布
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#数学分析
本文详细介绍了100个积分公式中的71至100，涵盖了含有tanx, cotx, secx, cscx, ex, lnx等形式的积分计算。每个公式都附有完整的证明过程，例如∫1±tanx1dx=21(x±ln|cosx±sinx|)+C等。通过这些公式，读者可以深入理解积分计算的方法和技巧。"
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索引
含有 tan x , cot x , sec x , csc x \tan x,\cot x,\sec x,\csc x tanx,cotx,secx,cscx的形式
71. ∫ 1 1 ± tan x d x = 1 2 ( x ± ln ∣ cos x ± sin x ∣ ) + C \int{\frac{1}{1\pm \tan x}dx=\frac{1}{2}\left( x\pm \ln \left| \cos x\pm \sin x \right| \right)+C} ∫1±tanx1dx=21(x±ln∣cosx±sinx∣)+C
72. ∫ 1 1 ± cot x d x = 1 2 ( x ∓ ln ∣ sin x ± cos x ∣ ) + C \int_{
{}}^{
{}}{\frac{1}{1\pm \cot x}dx}=\frac{1}{2}\left( x\mp \ln \left| \sin x\pm \cos x \right| \right)+C ∫1±cotx1dx=21(x∓ln∣sinx±cosx∣)+C
73. ∫ 1 1 ± sec x d x = x + cot x ∓ csc x + C \int_{
{}}^{
{}}{\frac{1}{1\pm \sec x}dx}=x+\cot x\mp \csc x+C ∫1±secx1dx=x+cotx∓cscx+C
74. ∫ 1 1 ± csc x d x = x − tan x ± sec x + C \int_{
{}}^{
{}}{\frac{1}{1\pm \csc x}dx}=x-\tan x\pm \sec x+C ∫1±cscx1dx=x−tanx±secx+C
75. ∫ arcsin x d x = x arcsin x + 1 − x 2 + C \int_{
{}}^{
{}}{\arcsin xdx}=x\arcsin x+\sqrt[{}]{1-{
{x}^{2}}}+C ∫arcsinxdx=xarcsinx+1−x2
+C
76. ∫ arccos x d x = x arccos x − 1 − x 2 + C \int_{
{}}^{
{}}{\arccos xdx}=x\arccos x-\sqrt[{}]{1-{
{x}^{2}}}+C ∫arccosxdx=xarccosx−1−x2
+C
77. ∫ arctan x d x = x arctan x − 1 2 ln ( 1 + x 2 ) + C \int_{
{}}^{
{}}{\arctan xdx}=x\arctan x-\frac{1}{2}\ln \left( 1+{
{x}^{2}} \right)+C ∫arctanxdx=xarctanx−21ln(1+x2)+C
78. ∫ a r c c o t x d x = x ⋅ a r c c o t x + 1 2 ln ( 1 + x 2 ) + C \int_{
{}}^{
{}}{arccot xdx}=x\centerdot arccotx+\frac{1}{2}\ln \left( 1+{
{x}^{2}} \right)+C ∫arccotxdx=x⋅arccotx+21ln(1+x2)+C
79. ∫ a r c s e c x d x = x ⋅ a r c s e c x − ln ∣ x + x 2 − 1 ∣ + C \int_{
{}}^{
{}}{arcsec xdx}=x\centerdot arcsec x-\ln \left| x+\sqrt[{}]{
{
{x}^{2}}-1} \right|+C ∫arcsecxdx=x⋅arcsecx−ln∣∣x+x2−1
∣∣+C
80. ∫ a r c c s c x d x = x ⋅ a r c c s c x + ln ∣ x + x 2 − 1 ∣ + C \int_{
{}}^{
{}}{arccsc xdx}=x\centerdot arccsc x+\ln \left| x+\sqrt[{}]{
{
{x}^{2}}-1} \right|+C ∫arccscxdx=x⋅arccscx+ln∣∣x+x2−1
∣∣+C
81. ∫ x arcsin x d x = 1 4 [ x 1 − x 2 + ( 2 x 2 − 1 ) arcsin x ] + C \int_{
{}}^{
{}}{x\arcsin xdx}=\frac{1}{4}\left[ x\sqrt[{}]{1-{
{x}^{2}}}+\left( 2{
{x}^{2}}-1 \right)\arcsin x \right]+C ∫xarcsinxdx=41[x1−x2
+(2x2−1)arcsinx]+C
82. ∫ x arccos x d x = 1 4 [ − x 1 − x 2 + ( 2 x 2 − 1 ) arccos x ] + C \int_{
{}}^{
{}}{x\arccos xdx}=\frac{1}{4}\left[ -x\sqrt[{}]{1-{
{x}^{2}}}+\left( 2{
{x}^{2}}-1 \right)\arccos x \right]+C ∫xarccosxdx=41[−x1−x2
+(2x2−1)arccosx]+C
83. ∫ x arctan x d x = 1 2 [ ( 1 + x 2 ) arctan x − x ] + C \int_{
{}}^{
{}}{x\arctan xdx}=\frac{1}{2}\left[ \left( 1+{
{x}^{2}} \right)\arctan x-x \right]+C ∫xarctanxdx=21[(1+x2)arctanx−x]+C
84. ∫ x ⋅ a r c c o t x d x = 1 2 [ ( 1 + x 2 ) a r c c o t x + x ] + C \int_{
{}}^{
{}}{x\centerdot arccot xdx}=\frac{1}{2}\left[ \left( 1+{
{x}^{2}} \right)arccot x+x \right]+C ∫x⋅arccotxdx=21[(1+x2)arccotx+x]+C
含有 e x {
{e}^{x}} ex的形式
85. ∫ a x d x = a x ln a + C \int_{
{}}^{
{}}{
{
{a}^{x}}dx}=\frac{
{
{a}^{x}}}{\ln a}+C ∫axdx=lnaax+C
86. ∫ e x d x = e x + C \int_{
{}}^{
{}}{
{
{e}^{x}}dx}={
{e}^{x}}+C ∫exdx=ex+C
87. ∫ x e x d x = ( x − 1 ) e x + C \int_{
{}}^{
{}}{x{
{e}^{x}}dx}=\left( x-1 \right){
{e}^{x}}+C ∫xexdx=(x−1)ex+C
88. ∫ x n e x d x = x n e x − n ∫ x n − 1 e x d x \int_{
{}}^{
{}}{
{
{x}^{n}}{
{e}^{x}}dx}={
{x}^{n}}{
{e}^{x}}-n\int_{
{}}^{
{}}{
{
{x}^{n-1}}{
{e}^{x}}dx} ∫xnexdx=xnex−n∫xn−1exdx
89. ∫ 1 1 + e x d x = x − ln ( 1 + e x ) + C \int_{
{}}^{
{}}{\frac{1}{1+{
{e}^{x}}}dx}=x-\ln \left( 1+{
{e}^{x}} \right)+C ∫1+ex1dx=x−ln(1+ex)+C
90. I 1 = ∫ e a x sin b x d x = e a x a 2 + b 2 ( a sin b x − b cos b x ) + C {
{I}_{1}}=\int_{
{}}^{
{}}{
{
{e}^{ax}}\sin bxdx}=\frac{
{
{e}^{ax}}}{
{
{a}^{2}}+{
{b}^{2}}}\left( a\sin bx-b\cos bx \right)+C I1=∫eaxsinbxdx=a2+b2eax(asinbx−bcosbx)+C
91. I 2 = ∫ e a x cos b x d x = e a x a 2 + b 2 ( a cos b x + b sin b x ) + C {
{I}_{2}}=\int_{
{}}^{
{}}{
{
{e}^{ax}}\cos bxdx}=\frac{
{
{e}^{ax}}}{
{
{a}^{2}}+{
{b}^{2}}}\left( a\cos bx+b\sin bx \right)+C I2=∫eaxcosbxdx=a2+b2eax(acosbx+bsinbx)+C
含有 ln x \ln x lnx的形式
92. ∫ ln x d x = x ( ln x − 1 ) + C \int_{
{}}^{
{}}{\ln xdx}=x\left( \ln x-1 \right)+C ∫lnxdx=x(lnx−1)+C
93. ∫ ln x x d x = 2 x ln x − 4 x + C \int_{
{}}^{
{}}{\frac{\ln x}{\sqrt[{}]{x}}dx}=2\sqrt[{}]{x}\ln x-4\sqrt[{}]{x}+C ∫x
lnxdx=2x
lnx−4x
+C
94. ∫ x ln x d x = x 2 4 ( 2 ln x − 1 ) + C \int_{
{}}^{
{}}{x\ln xdx}=\frac{
{
{x}^{2}}}{4}\left( 2\ln x-1 \right)+C ∫xlnxdx=4x2(2lnx−1)+C
95. ∫ x n ln x d x = x n + 1 ( n + 1 ) 2 [ ( n + 1 ) ln x − 1 ] + C , n ≠ − 1 \int_{
{}}^{
{}}{
{
{x}^{n}}\ln xdx}=\frac{
{
{x}^{n+1}}}{
{
{\left( n+1 \right)}^{2}}}\left[ \left( n+1 \right)\ln x-1 \right]+C,\text{ }n\ne -1 ∫xnlnxdx=(n+1)2xn+1[(n+1)lnx−1]+C, n=−1
96. ∫ ( ln x ) 2 d x = x [ ( ln x ) 2 − 2 ln x + 2 ] + C \int_{
{}}^{
{}}{
{
{\left( \ln x \right)}^{2}}dx}=x\left[ {
{\left( \ln x \right)}^{2}}-2\ln x+2 \right]+C ∫(lnx)2dx=x[(lnx)2−2lnx+2]+C
97. ∫ ( ln x ) n d x = x ( ln x ) n − n ∫ ( ln x ) n − 1 d x \int_{
{}}^{
{}}{
{
{\left( \ln x \right)}^{n}}dx}=x{
{\left( \ln x \right)}^{n}}-n\int_{
{}}^{
{}}{
{
{\left( \ln x \right)}^{n-1}}dx} ∫(lnx)ndx=x(lnx)n−n∫(lnx)n−1dx
98. ∫ sin ( ln x ) d x = x 2 [ sin ( ln x ) − cos ( ln x ) ] + C \int_{
{}}^{
{}}{\sin \left( \ln x \right)dx}=\frac{x}{2}\left[ \sin \left( \ln x \right)-\cos \left( \ln x \right) \right]+C ∫sin(lnx)dx=2x[sin(lnx)−cos(lnx)]+C
99. ∫ cos ( ln x ) d x = x 2 [ sin ( ln x ) + cos ( ln x ) ] + C \int_{
{}}^{
{}}{\cos \left( \ln x \right)dx}=\frac{x}{2}\left[ \sin \left( \ln x \right)+\cos \left( \ln x \right) \right]+C ∫cos(lnx)dx=2x[sin(lnx)+cos(lnx)]+C
100. ∫ ln ( x + 1 + x 2 ) d x = x ln ( x + 1 + x 2 ) − 1 + x 2 + C \int_{
{}}^{
{}}{\ln \left( x+\sqrt[{}]{1+{
{x}^{2}}} \right)dx}=x\ln \left( x+\sqrt[{}]{1+{
{x}^{2}}} \right)-\sqrt[{}]{1+{
{x}^{2}}}+C ∫ln(x+1+x2
)dx=xln(x+1+x2
)−1+x2
+C
含有 tan x , cot x , sec x , csc x \tan x,\cot x,\sec x,\csc x tanx,cotx,secx,cscx的形式
71. ∫ 1 1 ± tan x d x = 1 2 ( x ± ln ∣ cos x ± sin x ∣ ) + C \int{\frac{1}{1\pm \tan x}dx=\frac{1}{2}\left( x\pm \ln \left| \cos x\pm \sin x \right| \right)+C} ∫1±tanx1dx=21(x±ln∣cosx±sinx∣)+C
证明： ∫ 1 1 ± tan x d x t = tan x ‾ ‾ ∫ 1 1 ± t d ( arctan t ) = ∫ 1 1 ± t ⋅ 1 1 + t 2 d t = 1 2 ∫ ( 1 1 ± t + 1 ∓ t 1 + t 2 ) d t = 1 2 ∫ ( 1 1 ± t + 1 1 + t 2 ∓ t 1 + t 2 ) d t = 1 2 [ ± ln ∣ 1 ± t ∣ + arctan t ∓ 1 2 ln ( 1 + t 2 ) ] + C = 1 2 [ ± ln ∣ 1 ± tan x ∣ + x ∓ ln 1 + tan 2 x ] + C = 1 2 ( x ± ln ∣ 1 ± tan x sec x ∣ ) + C = 1 2 ( x ± ln ∣ cos x ± sin x ∣ ) + C \begin{aligned} & \int{\frac{1}{1\pm \tan x}dx}\text{ }\underline{\underline{t=\tan x}}\text{ }\int{\frac{1}{1\pm t}d\left( \arctan t \right)} \\ & =\int{\frac{1}{1\pm t}\centerdot \frac{1}{1+{
{t}^{2}}}dt} \\ & =\frac{1}{2}\int{\left( \frac{1}{1\pm t}+\frac{1\mp t}{1+{
{t}^{2}}} \right)dt} \\ & =\frac{1}{2}\int{\left( \frac{1}{1\pm t}+\frac{1}{1+{
{t}^{2}}}\mp \frac{t}{1+{
{t}^{2}}} \right)}dt \\ & =\frac{1}{2}\left[ \pm \ln \left| 1\pm t \right|+\arctan t\mp \frac{1}{2}\ln \left( 1+{
{t}^{2}} \right) \right]+C \\ & =\frac{1}{2}\left[ \pm \ln \left| 1\pm \tan x \right|+x\mp \ln \sqrt[{}]{1+{
{\tan }^{2}}x} \right]+C \\ & =\frac{1}{2}\left( x\pm \ln \left| \frac{1\pm \tan x}{\sec x} \right| \right)+C \\ & =\frac{1}{2}\left( x\pm \ln \left| \cos x\pm \sin x \right| \right)+C \\ \end{aligned} ∫1±tanx1dx t=tanx ∫1±t1d(arctant)=∫1±t1⋅1+t21dt=21∫(1±t1+1+t21∓t)dt=21∫(1±t1+1+t21∓1+t2t)dt=21[±ln∣1±t∣+arctant∓21ln(1+t2)]+C=21[±ln∣1±tanx∣+x∓ln1+tan2x
]+C=21(x±ln∣∣∣∣secx1±tanx∣∣∣∣)+C=21(x±ln∣cosx±sinx∣)+C
72. ∫ 1 1 ± cot x d x = 1 2 ( x ∓ ln ∣ sin x ± cos x ∣ ) + C \int_{
{}}^{
{}}{\frac{1}{1\pm \cot x}dx}=\frac{1}{2}\left( x\mp \ln \left| \sin x\pm \cos x \right| \right)+C ∫1±cotx1dx=21(x∓ln∣sinx±cosx∣)+C
证明： ∫ 1 1 ± cot x d x t = cot x ‾ ‾ ∫ 1 1 ± t d ( a r c c o t t ) = − ∫ 1 1 ± t ⋅ 1 1 + t 2 d t = − 1 2 ∫ ( 1 1 ± t + 1 1 + t 2 ∓ t 1 + t 2 ) d t = − 1 2 [ ± ln ∣ 1 ± t ∣ − a r c c o t t ∓ 1 2 ln ( 1 + t 2 ) ] + C = − 1 2 [ ± ln ∣ 1 ± cot x ∣ − x ∓ ln 1 + cot 2 x ] + C = 1 2 [ x ± ln 1 + cot 2 x ∓ ln ∣ 1 ± cot x ∣ ] + C = 1 2 [ x ± ln ∣ csc x 1 ± cot x ∣ ] + C = 1 2 [ x ∓ ln ∣ sin x ± cos x ∣ ] + C \begin{aligned} & \int{\frac{1}{1\pm \cot x}dx\text{ }}\underline{\underline{t=\cot x}}\text{ }\int{\frac{1}{1\pm t}d\left( arccot t \right)} \\ & =-\int{\frac{1}{1\pm t}}\centerdot \frac{1}{1+{
{t}^{2}}}dt \\ & =-\frac{1}{2}\int{\left( \frac{1}{1\pm t}+\frac{1}{1+{
{t}^{2}}}\mp \frac{t}{1+{
{t}^{2}}} \right)dt} \\ & =-\frac{1}{2}\left[ \pm \ln \left| 1\pm t \right|-arccot t\mp \frac{1}{2}\ln \left( 1+{
{t}^{2}} \right) \right]+C \\ & =-\frac{1}{2}\left[ \pm \ln \left| 1\pm \cot x \right|-x\mp \ln \sqrt[{}]{1+{
{\cot }^{2}}x} \right]+C \\ & =\frac{1}{2}\left[ x\pm \ln \sqrt[{}]{1+{
{\cot }^{2}}x}\mp \ln \left| 1\pm \cot x \right| \right]+C \\ & =\frac{1}{2}\left[ x\pm \ln \left| \frac{\csc x}{1\pm \cot x} \right| \right]+C \\ & =\frac{1}{2}\left[ x\mp \ln \left| \sin x\pm \cos x \right| \right]+C \\ \end{aligned} ∫1±cotx1dx t=cotx ∫1±t1d(arccott)=−∫1±t1⋅1+t21dt=−21∫(1±t1+1+t21∓1+t2t)dt=−21[±ln∣1±t∣−arccott∓21ln(1+t2)]+C=−21[±ln∣1±cotx∣−x∓ln1+cot2x
]+C=21[x±ln1+cot2x
∓ln∣1±cotx∣]+C=21[x±ln∣∣∣∣1±cotxcscx∣∣∣∣]+C=21[x∓ln∣sinx±cosx∣]+C
73. ∫ 1 1 ± sec x d x = x + cot x ∓ csc x + C \int_{
{}}^{
{}}{\frac{1}{1\pm \sec x}dx}=x+\cot x\mp \csc x+C ∫1±secx1dx=x+cotx∓cscx+C
证明： ∫ 1 1 ± sec x d x = ∫ 1 1 ± 1 cos x d x = ∫ cos x cos x ± 1 d x = ∫ d x ∓ ∫ 1 cos x ± 1 d x = x ± ∫ cos x ∓ 1 sin 2 x d x = x ± ∫ ( csc x cot x ∓ csc 2 x ) d x = x ± ∫ d ( − csc x ± cot x ) = x ± ( − csc x ± cot x ) + C = x + cot x ∓ csc x + C \begin{aligned} & \int_{
{}}^{
{}}{\frac{1}{1\pm \sec x}dx}=\int_{
{}}^{
{}}{\frac{1}{1\pm \frac{1}{\cos x}}dx} \\ & =\int_{
{}}^{
{}}{\frac{\cos x}{\cos x\pm 1}dx} \\ & =\int_{
{}}^{
{}}{dx}\mp \int_{
{}}^{
{}}{\frac{1}{\cos x\pm 1}dx} \\ & =x\pm \int_{
{}}^{
{}}{\frac{\cos x\mp 1}{
{
{\sin }^{2}}x}dx} \\ & =x\pm \int_{
{}}^{
{}}{\left( \csc x\cot x\mp {
{\csc }^{2}}x \right)dx} \\ & =x\pm \int_{
{}}^{
{}}{d\left( -\csc x\pm \cot x \right)} \\ & =x\pm \left( -\csc x\pm \cot x \right)+C \\ & =x+\cot x\mp \csc x+C \\ \end{aligned} ∫1±secx1dx=∫1±cosx11dx=∫cosx±1cosxdx=∫dx∓∫cosx±11dx=x±∫sin2xcosx∓1dx=x±∫(cscxcotx∓csc2x)dx=x±∫d(−cscx±cotx)=x±(−cscx±cotx)+C=x+cotx∓cscx+C
74. ∫ 1 1 ± csc x d x = x − tan x ± sec x + C \int_{
{}}^{
{}}{\frac{1}{1\pm \csc x}dx}=x-\tan x\pm \sec x+C ∫1±cscx1dx=x−tanx±secx+C
证明： ∫ 1 1 ± csc x d x = ∫ 1 1 ± 1 sin x d x = ∫ sin x sin x ± 1 d x = ∫ d x ± ∫ sin x ∓ 1 cos 2 x d x = x ± ∫ ( sec x tan x ∓ sec 2 x ) d x = x ± ∫ d ( sec x ∓ tan x ) = x ± ( sec x ∓ tan x ) + C = x − tan x ± sec x + C \begin{aligned} & \int_{
{}}^{
{}}{\frac{1}{1\pm \csc x}dx}=\int_{
{}}^{
{}}{\frac{1}{1\pm \frac{1}{\sin x}}dx} \\ & =\int_{
{}}^{
{}}{\frac{\sin x}{\sin x\pm 1}dx} \\ & =\int_{
{}}^{
{}}{dx}\pm \int_{
{}}^{
{}}{\frac{\sin x\mp 1}{
{
{\cos }^{2}}x}dx} \\ & =x\pm \int_{
{}}^{
{}}{\left( \sec x\tan x\mp {
{\sec }^{2}}x \right)dx} \\ & =x\pm \int_{
{}}^{
{}}{d\left( \sec x\mp \tan x \right)} \\ & =x\pm \left( \sec x\mp \tan x \right)+C \\ & =x-\tan x\pm \sec x+C \\ \end{aligned} ∫1±cscx1dx=∫1±sinx11dx=∫sinx±1sinxdx=∫dx±∫cos2xsinx∓1dx=x±∫(secxtanx∓sec2x)dx=x±∫d(secx∓tanx)=x±(secx∓tanx)+C=x−tanx±secx+C
75. ∫ arcsin x d x = x arcsin x + 1 − x 2 + C \int_{
{}}^{
{}}{\arcsin xdx}=x\arcsin x+\sqrt[{}]{1-{
{x}^{2}}}+C ∫arcsinxdx=xarcsinx+1−x2
+C
证明： ∫ arcsin x d x = x arcsin x − ∫ x d ( arcsin x ) = x arcsin x − ∫ x 1 − x 2 d x = x arcsin x + ∫ 1 2 1 − x 2 d ( 1 − x 2 ) = x arcsin x + 1 − x 2 + C \begin{aligned} & \int_{
{}}^{
{}}{\arcsin xdx}=x\arcsin x-\int_{
{}}^{
{}}{xd\left( \arcsin x \right)} \\ & =x\arcsin x-\int_{
{}}^{
{}}{\frac{x}{\sqrt[{}]{1-{
{x}^{2}}}}dx} \\ & =x\arcsin x+\int_{
{}}^{
{}}{\frac{1}{2\sqrt[{}]{1-{
{x}^{2}}}}d\left( 1-{
{x}^{2}} \right)} \\ & =x\arcsin x+\sqrt[{}]{1-{
{x}^{2}}}+C \\ \end{aligned} ∫arcsinxdx=xarcsinx−∫xd(arcsinx)=xarcsinx−∫1−x2
xdx=xarcsinx+∫21−x2
1d(1−x2)=xarcsinx+1−x2
+C
76. ∫ arccos x d x = x arccos x − 1 − x 2 + C \int_{
{}}^{
{}}{\arccos xdx}=x\arccos x-\sqrt[{}]{1-{
{x}^{2}}}+C ∫arccosxdx=xarccosx−1−x2
+C
证明： ∫ arccos x d x = x arccos x − ∫ x d ( arccos x ) = x arccos x + ∫ x 1 − x 2 d x = x arccos x − ∫ 1 2 1 − x 2 d ( 1 − x 2 ) = x arccos x − 1 − x 2 + C \begin{aligned} & \int_{
{}}^{
{}}{\arccos xdx}=x\arccos x-\int_{
{}}^{
{}}{xd\left( \arccos x \right)} \\ & =x\arccos x+\int_{
{}}^{
{}}{\frac{x}{\sqrt[{}]{1-{
{x}^{2}}}}dx} \\ & =x\arccos x-\int_{
{}}^{
{}}{\frac{1}{2\sqrt[{}]{1-{
{x}^{2}}}}d\left( 1-{
{x}^{2}} \right)} \\ & =x\arccos x-\sqrt[{}]{1-{
{x}^{2}}}+C \\ \end{aligned} ∫arccosxdx=xarccosx−∫xd(arccosx)=xarccosx+∫1−x2
xdx=xarccosx−∫21−x2
1d(1−x2)=xarccosx−1−x2
+C
77. ∫ arctan x d x = x arctan x − 1 2 ln ( 1 + x 2 ) + C \int_{
{}}^{
{}}{\arctan xdx}=x\arctan x-\frac{1}{2}\ln \left( 1+{
{x}^{2}} \right)+C ∫arctanxdx=xarctanx−21ln(1+x2)+C
证明： ∫ arctan x d x = x arctan x − ∫ x d ( arctan x ) = x arctan x − ∫ x 1 + x 2 d x = x arctan x − 1 2 ∫ 1 1 + x 2 d ( 1 + x 2 ) = x arctan x − 1 2 ln ( 1 + x 2 ) + C \begin{aligned} & \int_{
{}}^{
{}}{\arctan xdx}=x\arctan x-\int_{
{}}^{
{}}{xd\left( \arctan x \right)} \\ & =x\arctan x-\int_{
{}}^{
{}}{\frac{x}{1+{
{x}^{2}}}dx} \\ & =x\arctan x-\frac{1}{2}\int_{
{}}^{
{}}{\frac{1}{1+{
{x}^{2}}}d\left( 1+{
{x}^{2}} \right)} \\ & =x\arctan x-\frac{1}{2}\ln \left( 1+{
{x}^{2}} \right)+C \\ \end{aligned} ∫arctanxdx=xarctanx−∫