���;��G;$rrR�sb��f9���~����ߓs�O͹�o�'�џ��O�H΅��N��|�����I浙��v)tLb>�^X��֯V���ؒ ��0[�_9��{9���xk�b��>#igwG�5V����w��m?K�Ļ���V�7�]�0M ����Ę��$�˙��J�s��غ��8��JS�br� L'�n��^��-W�f������fQQ{� ,ܓc�°��9�#'���؎�WTf��~\{��a�:���cK. ʃ a dx = ax + k ʃ … (1) ∫dxa+b sin⁡2x\int \frac{dx}{a+b\: \sin ^{2}x}∫a+bsin2xdx​  or ∫dxa+b cos⁡2x\int \frac{dx}{a+b\: \cos ^{2}x}∫a+bcos2xdx​  or ∫dxasin⁡2x+bsin⁡xcos⁡x+ccos⁡2x\int \frac{dx}{a \sin ^{2}x+b \sin x\cos x+c\cos ^{2}x}∫asin2x+bsinxcosx+ccos2xdx​  put tan x = t. (2) ∫dxa+bsin⁡x\int \frac{dx}{a+b\sin x}∫a+bsinxdx​  or ∫dxa+bcos⁡x\int \frac{dx}{a+b\cos x}∫a+bcosxdx​  or. (i) ∫ (ax+b)n dx = (ax+b)n+1a(n+1) +c\frac{(ax+b)^{n+1}}{a(n+1)} \: + ca(n+1)(ax+b)n+1​+c , n ≠ -1, (ii) ∫dxax+b=1a\int \frac{dx}{ax+b} = \frac{1}{a}∫ax+bdx​=a1​ ln(ax+b) +c, (iv) ∫ apx+q dx = 1papx+qln⁡a +c\frac{1}{p} \frac{a^{px+q}}{\ln a} \: +cp1​lnaapx+q​+c , a>0, (v)  ∫ sin (ax+b)dx = -(1/a) cos (ax+b) + c, (vi) ∫ cos (ax+b) dx = (1/a) sin (ax+b) + c, (vii)  ∫ tan (ax+b)dx = (1/a) ln sec (ax+b) + c, (viii) ∫ cot (ax+b)dx = (1/a) ln sin (ax+b) + c, (ix) ∫ sec2 (ax+b)dx = (1/a) tan (ax+b) + c, (x) ∫ cosec2 (ax+b) dx = -(1/a) cot (ax+b) + c, (xi) ∫ sec (ax+b) ⋅ tan (ax+b) dx = (1/a) sec (ax+b) + c, (xii) ∫ cosec (ax+b) ⋅ cot (ax+b) dx = -(1/a) cosec (ax+b) + c, (xiii) ∫ sec x dx = ln (sec x + tan x) + c  or  ln⁡tan(π4+x2)\ln tan \left ( \frac{\pi }{4} +\frac{x}{2}\right )lntan(4π​+2x​), (xiv) ∫ cosec x dx = ln (cosec x – cot x) + c  or  ln tan (x/2) +c  or –ln (cosec x + cot x)+ c, (xv)  ∫dxa2−x2=sin⁡−1xa+c\int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin ^{-1} \frac{x}{a} +c∫a2−x2​dx​=sin−1ax​+c, (xvi) ∫dxa2+x2=12tan⁡−1xa+c\int \frac{dx}{a^{2}+x^{2}} = \frac{1}{2} \tan ^{-1} \frac{x}{a} +c∫a2+x2dx​=21​tan−1ax​+c, (xvii) ∫dx∣x∣x2−a2=1asec⁡−1xa\int \frac{dx}{\left | x \right |\sqrt{x^{2}-a^{2}}} = \frac{1}{a} \sec ^{-1}\frac{x}{a}∫∣x∣x2−a2​dx​=a1​sec−1ax​, (xviii) ∫dxx2+a2=ln⁡[x+x2+a2]+c\int \frac{dx}{\sqrt{x^{2}+a^{2}}} = \ln [x+\sqrt{x^{2}+a^{2}}]+c∫x2+a2​dx​=ln[x+x2+a2​]+c, (xix) ∫dxx2−a2=ln⁡[x+x2−a2]+c\int \frac{dx}{\sqrt{x^{2}-a^{2}}} = \ln [x+\sqrt{x^{2}-a^{2}}]+c∫x2−a2​dx​=ln[x+x2−a2​]+c, (xx) ∫dxa2−x2=12aln⁡∣a+xa−x∣+c\int \frac{dx}{{a^{2}-x^{2}}} = \frac{1}{2a}\ln \left | \frac{a+x}{a-x} \right |+c∫a2−x2dx​=2a1​ln∣∣∣​a−xa+x​∣∣∣​+c, (xxi) ∫dxx2−a2=12aln⁡∣x−ax+a∣+c\int \frac{dx}{{x^{2}-a^{2}}} = \frac{1}{2a}\ln \left | \frac{x-a}{x+a} \right |+c∫x2−a2dx​=2a1​ln∣∣∣​x+ax−a​∣∣∣​+c, (xxii) ∫a2−x2 dx=x2a2−x2+a22sin⁡−1xa+c\int \sqrt{a^{2}-x^{2}} \: dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin ^{-1}\frac{x}{a}+c∫a2−x2​dx=2x​a2−x2​+2a2​sin−1ax​+c, (xxiii) ∫x2+a2 dx=x2x2+a2+a22ln⁡(x+x2+a2a)+c\int \sqrt{x^{2}+a^{2}} \: dx = \frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\ln \left ( \frac{x+\sqrt{x^{2}+a^{2}}}{a} \right )+c∫x2+a2​dx=2x​x2+a2​+2a2​ln(ax+x2+a2​​)+c, (xxiv) ∫x2−a2 dx=x2x2−a2−a22ln⁡(x+x2−a2a)+c\int \sqrt{x^{2}-a^{2}} \: dx = \frac{x}{2}\sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2}\ln \left ( \frac{x+\sqrt{x^{2}-a^{2}}}{a} \right )+c∫x2−a2​dx=2x​x2−a2​−2a2​ln(ax+x2−a2​​)+c, (xxv) ∫eax.sin⁡bx dx=eaxa2+b2(a sin⁡bx−bcos⁡bx)+c\int e^{ax}.\sin bx \: dx = \frac{e^{ax}}{a^{2}+b^{2}}\left ( a\, \sin bx-b\cos bx \right )+c∫eax.sinbxdx=a2+b2eax​(asinbx−bcosbx)+c, (xxvi) ∫eax.cos⁡bx dx=eaxa2+b2(a cos⁡bx+b sin⁡bx)+c\int e^{ax}.\cos bx \: dx = \frac{e^{ax}}{a^{2}+b^{2}}\left ( a\, \cos bx+b\: \sin bx \right )+c∫eax.cosbxdx=a2+b2eax​(acosbx+bsinbx)+c, (ii) ∫(f(x)±g(x)) dx=∫f(x) dx±g(x) dx\int (f(x)\pm g(x))\: dx = \int f(x)\: dx\pm g(x)\: dx∫(f(x)±g(x))dx=∫f(x)dx±g(x)dx, (iii) ∫ f(x) dx =  g(x)+c ⇒ ∫ f(ax+b) dx = g(ax+b)a+c\frac{g(ax+b)}{a}+cag(ax+b)​+c, If we substitute f(x) = t, then f’(x) dx = dt, ∫( f(x) g(x)) dx = f(x) ∫g(x) dx – ∫(ddx(f(x))∫(g(x))dx)dx\int\left ( \frac{d}{dx}(f(x))\int (g(x))dx\right ) dx∫(dxd​(f(x))∫(g(x))dx)dx. Excel File Stops Scrolling, Epitaph Meaning In Urdu, Dragon Ball Z - Attack Of The Saiyans Emulator, Awaken Meaning In Tamil, 1870 Presidential Election, 1 To 10 In English Spelling, Nyc Doe Ipad Unlock, Voulez-vous Danser Translation, Landscape Mode In Mobile Camera, Richard Brunelle Cause Of Death, Epitaph Meaning In Urdu, Blog Introduction Samples, " /> ���;��G;$rrR�sb��f9���~����ߓs�O͹�o�'�џ��O�H΅��N��|�����I浙��v)tLb>�^X��֯V���ؒ ��0[�_9��{9���xk�b��>#igwG�5V����w��m?K�Ļ���V�7�]�0M ����Ę��$�˙��J�s��غ��8��JS�br� L'�n��^��-W�f������fQQ{� ,ܓc�°��9�#'���؎�WTf��~\{��a�:���cK. ʃ a dx = ax + k ʃ … (1) ∫dxa+b sin⁡2x\int \frac{dx}{a+b\: \sin ^{2}x}∫a+bsin2xdx​  or ∫dxa+b cos⁡2x\int \frac{dx}{a+b\: \cos ^{2}x}∫a+bcos2xdx​  or ∫dxasin⁡2x+bsin⁡xcos⁡x+ccos⁡2x\int \frac{dx}{a \sin ^{2}x+b \sin x\cos x+c\cos ^{2}x}∫asin2x+bsinxcosx+ccos2xdx​  put tan x = t. (2) ∫dxa+bsin⁡x\int \frac{dx}{a+b\sin x}∫a+bsinxdx​  or ∫dxa+bcos⁡x\int \frac{dx}{a+b\cos x}∫a+bcosxdx​  or. (i) ∫ (ax+b)n dx = (ax+b)n+1a(n+1) +c\frac{(ax+b)^{n+1}}{a(n+1)} \: + ca(n+1)(ax+b)n+1​+c , n ≠ -1, (ii) ∫dxax+b=1a\int \frac{dx}{ax+b} = \frac{1}{a}∫ax+bdx​=a1​ ln(ax+b) +c, (iv) ∫ apx+q dx = 1papx+qln⁡a +c\frac{1}{p} \frac{a^{px+q}}{\ln a} \: +cp1​lnaapx+q​+c , a>0, (v)  ∫ sin (ax+b)dx = -(1/a) cos (ax+b) + c, (vi) ∫ cos (ax+b) dx = (1/a) sin (ax+b) + c, (vii)  ∫ tan (ax+b)dx = (1/a) ln sec (ax+b) + c, (viii) ∫ cot (ax+b)dx = (1/a) ln sin (ax+b) + c, (ix) ∫ sec2 (ax+b)dx = (1/a) tan (ax+b) + c, (x) ∫ cosec2 (ax+b) dx = -(1/a) cot (ax+b) + c, (xi) ∫ sec (ax+b) ⋅ tan (ax+b) dx = (1/a) sec (ax+b) + c, (xii) ∫ cosec (ax+b) ⋅ cot (ax+b) dx = -(1/a) cosec (ax+b) + c, (xiii) ∫ sec x dx = ln (sec x + tan x) + c  or  ln⁡tan(π4+x2)\ln tan \left ( \frac{\pi }{4} +\frac{x}{2}\right )lntan(4π​+2x​), (xiv) ∫ cosec x dx = ln (cosec x – cot x) + c  or  ln tan (x/2) +c  or –ln (cosec x + cot x)+ c, (xv)  ∫dxa2−x2=sin⁡−1xa+c\int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin ^{-1} \frac{x}{a} +c∫a2−x2​dx​=sin−1ax​+c, (xvi) ∫dxa2+x2=12tan⁡−1xa+c\int \frac{dx}{a^{2}+x^{2}} = \frac{1}{2} \tan ^{-1} \frac{x}{a} +c∫a2+x2dx​=21​tan−1ax​+c, (xvii) ∫dx∣x∣x2−a2=1asec⁡−1xa\int \frac{dx}{\left | x \right |\sqrt{x^{2}-a^{2}}} = \frac{1}{a} \sec ^{-1}\frac{x}{a}∫∣x∣x2−a2​dx​=a1​sec−1ax​, (xviii) ∫dxx2+a2=ln⁡[x+x2+a2]+c\int \frac{dx}{\sqrt{x^{2}+a^{2}}} = \ln [x+\sqrt{x^{2}+a^{2}}]+c∫x2+a2​dx​=ln[x+x2+a2​]+c, (xix) ∫dxx2−a2=ln⁡[x+x2−a2]+c\int \frac{dx}{\sqrt{x^{2}-a^{2}}} = \ln [x+\sqrt{x^{2}-a^{2}}]+c∫x2−a2​dx​=ln[x+x2−a2​]+c, (xx) ∫dxa2−x2=12aln⁡∣a+xa−x∣+c\int \frac{dx}{{a^{2}-x^{2}}} = \frac{1}{2a}\ln \left | \frac{a+x}{a-x} \right |+c∫a2−x2dx​=2a1​ln∣∣∣​a−xa+x​∣∣∣​+c, (xxi) ∫dxx2−a2=12aln⁡∣x−ax+a∣+c\int \frac{dx}{{x^{2}-a^{2}}} = \frac{1}{2a}\ln \left | \frac{x-a}{x+a} \right |+c∫x2−a2dx​=2a1​ln∣∣∣​x+ax−a​∣∣∣​+c, (xxii) ∫a2−x2 dx=x2a2−x2+a22sin⁡−1xa+c\int \sqrt{a^{2}-x^{2}} \: dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin ^{-1}\frac{x}{a}+c∫a2−x2​dx=2x​a2−x2​+2a2​sin−1ax​+c, (xxiii) ∫x2+a2 dx=x2x2+a2+a22ln⁡(x+x2+a2a)+c\int \sqrt{x^{2}+a^{2}} \: dx = \frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\ln \left ( \frac{x+\sqrt{x^{2}+a^{2}}}{a} \right )+c∫x2+a2​dx=2x​x2+a2​+2a2​ln(ax+x2+a2​​)+c, (xxiv) ∫x2−a2 dx=x2x2−a2−a22ln⁡(x+x2−a2a)+c\int \sqrt{x^{2}-a^{2}} \: dx = \frac{x}{2}\sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2}\ln \left ( \frac{x+\sqrt{x^{2}-a^{2}}}{a} \right )+c∫x2−a2​dx=2x​x2−a2​−2a2​ln(ax+x2−a2​​)+c, (xxv) ∫eax.sin⁡bx dx=eaxa2+b2(a sin⁡bx−bcos⁡bx)+c\int e^{ax}.\sin bx \: dx = \frac{e^{ax}}{a^{2}+b^{2}}\left ( a\, \sin bx-b\cos bx \right )+c∫eax.sinbxdx=a2+b2eax​(asinbx−bcosbx)+c, (xxvi) ∫eax.cos⁡bx dx=eaxa2+b2(a cos⁡bx+b sin⁡bx)+c\int e^{ax}.\cos bx \: dx = \frac{e^{ax}}{a^{2}+b^{2}}\left ( a\, \cos bx+b\: \sin bx \right )+c∫eax.cosbxdx=a2+b2eax​(acosbx+bsinbx)+c, (ii) ∫(f(x)±g(x)) dx=∫f(x) dx±g(x) dx\int (f(x)\pm g(x))\: dx = \int f(x)\: dx\pm g(x)\: dx∫(f(x)±g(x))dx=∫f(x)dx±g(x)dx, (iii) ∫ f(x) dx =  g(x)+c ⇒ ∫ f(ax+b) dx = g(ax+b)a+c\frac{g(ax+b)}{a}+cag(ax+b)​+c, If we substitute f(x) = t, then f’(x) dx = dt, ∫( f(x) g(x)) dx = f(x) ∫g(x) dx – ∫(ddx(f(x))∫(g(x))dx)dx\int\left ( \frac{d}{dx}(f(x))\int (g(x))dx\right ) dx∫(dxd​(f(x))∫(g(x))dx)dx. 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x in the terms of sines and cosines of multiples of x by using trigonometric results or De’ Moivere’s theorem. If P is a positive integer then integrand is expanded by binomial expansion. Z dx x = ln|x|+C 4. Where Pm-1 (x) is polynomial of degree (m-1) and k is constant. g. Integration by Parts. /Length 2560 Check the formula sheet of integration. Z cosxdx = sinx+C /Filter /FlateDecode INTEGRATION FORMULAS PDF. MyNotesAdda.com is an online Educational Platform, where you can download free PDF for UPSC, SSC CGL, BANK, RAILWAYS, RRB NTPC, LIC AAO, and many other exams. Z ex dx = ex +C 5. Case 2. Then it is made proper rational function by simple division and then partial fraction is used. Basic Integration Formulas 1. ∫ f(x)dx = g(x)+c  ⇔ ddx{g(x)+c}=f(x)\frac{d}{dx}\left \{g(x)+c \right \} = f(x)dxd​{g(x)+c}=f(x) , where c is called the constant of integration. If m and n are even natural numbers then express sin. ∫dxa+bsin⁡x+ccos⁡x\int \frac{dx}{a+b\sin x+c \cos x}∫a+bsinx+ccosxdx​  put tan (x/2) = t. (3) ∫acos⁡x+bsin⁡x+clcos⁡x+msin⁡x+ndx\int \frac{a\cos x+ b\sin x+c}{l\cos x+ m\sin x+n}dx∫lcosx+msinx+nacosx+bsinx+c​dx. x��\YsE~�_я�0][�DX^���],G �y�E�2�0f�fUWuWOg_3-ax�Zө���/��:�ǧ���� In general the funcyion as u is taken which comes first in the world ILATE. In general, integration is the reverse operation of differentiation. >> a. Φ‘ (z)} dz                     ;   Substitute   x =Φ(z), If u and v be two functions of x then  ∫ uv dx = u ∫ v dx – ​$$\int (\frac{du}{dx}\int v dx)dx$$​ is. Today, we are sharing INTEGRATION FORMULAS PDFi. Case 1. highest power of x in Q(x) < P(x)). (i) when you find integral ∫g(x) dx then it will not contain an arbitrary constant. (iv) ∫nta+nTf(x)dx=∫0af(x)dx,\int_{nt}^{a+nT}f(x)dx = \int_{0}^{a}f(x)dx ,∫nta+nT​f(x)dx=∫0a​f(x)dx,, n∈Z, a∈R . �*G�e���\GQ��p*rE{!��Jؔ�?��Jj�a ?MƇ�rm��ƇCt�7�"���4aEn�^���Q�xL�*/Ԭ���$�e��++ ��K)���=�a��}@GZ��gﯱ1�憚MHF��d�u�����X����hĉ;幟"&Fܧq_)�!V���,�ě�%�D+��^���~�����������X�La�X���S�[�,�o����l�>���;��G;$rrR�sb��f9���~����ߓs�O͹�o�'�џ��O�H΅��N��|�����I浙��v)tLb>�^X��֯V���ؒ ��0[�_9��{9���xk�b��>#igwG�5V����w��m?K�Ļ���V�7�]�0M ����Ę��\$�˙��J�s��غ��8��JS�br� L'�n��^��-W�f������fQQ{� ,ܓc�°��9�#'���؎�WTf��~\{��a�:���cK. ʃ a dx = ax + k ʃ … (1) ∫dxa+b sin⁡2x\int \frac{dx}{a+b\: \sin ^{2}x}∫a+bsin2xdx​  or ∫dxa+b cos⁡2x\int \frac{dx}{a+b\: \cos ^{2}x}∫a+bcos2xdx​  or ∫dxasin⁡2x+bsin⁡xcos⁡x+ccos⁡2x\int \frac{dx}{a \sin ^{2}x+b \sin x\cos x+c\cos ^{2}x}∫asin2x+bsinxcosx+ccos2xdx​  put tan x = t. (2) ∫dxa+bsin⁡x\int \frac{dx}{a+b\sin x}∫a+bsinxdx​  or ∫dxa+bcos⁡x\int \frac{dx}{a+b\cos x}∫a+bcosxdx​  or. (i) ∫ (ax+b)n dx = (ax+b)n+1a(n+1) +c\frac{(ax+b)^{n+1}}{a(n+1)} \: + ca(n+1)(ax+b)n+1​+c , n ≠ -1, (ii) ∫dxax+b=1a\int \frac{dx}{ax+b} = \frac{1}{a}∫ax+bdx​=a1​ ln(ax+b) +c, (iv) ∫ apx+q dx = 1papx+qln⁡a +c\frac{1}{p} \frac{a^{px+q}}{\ln a} \: +cp1​lnaapx+q​+c , a>0, (v)  ∫ sin (ax+b)dx = -(1/a) cos (ax+b) + c, (vi) ∫ cos (ax+b) dx = (1/a) sin (ax+b) + c, (vii)  ∫ tan (ax+b)dx = (1/a) ln sec (ax+b) + c, (viii) ∫ cot (ax+b)dx = (1/a) ln sin (ax+b) + c, (ix) ∫ sec2 (ax+b)dx = (1/a) tan (ax+b) + c, (x) ∫ cosec2 (ax+b) dx = -(1/a) cot (ax+b) + c, (xi) ∫ sec (ax+b) ⋅ tan (ax+b) dx = (1/a) sec (ax+b) + c, (xii) ∫ cosec (ax+b) ⋅ cot (ax+b) dx = -(1/a) cosec (ax+b) + c, (xiii) ∫ sec x dx = ln (sec x + tan x) + c  or  ln⁡tan(π4+x2)\ln tan \left ( \frac{\pi }{4} +\frac{x}{2}\right )lntan(4π​+2x​), (xiv) ∫ cosec x dx = ln (cosec x – cot x) + c  or  ln tan (x/2) +c  or –ln (cosec x + cot x)+ c, (xv)  ∫dxa2−x2=sin⁡−1xa+c\int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin ^{-1} \frac{x}{a} +c∫a2−x2​dx​=sin−1ax​+c, (xvi) ∫dxa2+x2=12tan⁡−1xa+c\int \frac{dx}{a^{2}+x^{2}} = \frac{1}{2} \tan ^{-1} \frac{x}{a} +c∫a2+x2dx​=21​tan−1ax​+c, (xvii) ∫dx∣x∣x2−a2=1asec⁡−1xa\int \frac{dx}{\left | x \right |\sqrt{x^{2}-a^{2}}} = \frac{1}{a} \sec ^{-1}\frac{x}{a}∫∣x∣x2−a2​dx​=a1​sec−1ax​, (xviii) ∫dxx2+a2=ln⁡[x+x2+a2]+c\int \frac{dx}{\sqrt{x^{2}+a^{2}}} = \ln [x+\sqrt{x^{2}+a^{2}}]+c∫x2+a2​dx​=ln[x+x2+a2​]+c, (xix) ∫dxx2−a2=ln⁡[x+x2−a2]+c\int \frac{dx}{\sqrt{x^{2}-a^{2}}} = \ln [x+\sqrt{x^{2}-a^{2}}]+c∫x2−a2​dx​=ln[x+x2−a2​]+c, (xx) ∫dxa2−x2=12aln⁡∣a+xa−x∣+c\int \frac{dx}{{a^{2}-x^{2}}} = \frac{1}{2a}\ln \left | \frac{a+x}{a-x} \right |+c∫a2−x2dx​=2a1​ln∣∣∣​a−xa+x​∣∣∣​+c, (xxi) ∫dxx2−a2=12aln⁡∣x−ax+a∣+c\int \frac{dx}{{x^{2}-a^{2}}} = \frac{1}{2a}\ln \left | \frac{x-a}{x+a} \right |+c∫x2−a2dx​=2a1​ln∣∣∣​x+ax−a​∣∣∣​+c, (xxii) ∫a2−x2 dx=x2a2−x2+a22sin⁡−1xa+c\int \sqrt{a^{2}-x^{2}} \: dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin ^{-1}\frac{x}{a}+c∫a2−x2​dx=2x​a2−x2​+2a2​sin−1ax​+c, (xxiii) ∫x2+a2 dx=x2x2+a2+a22ln⁡(x+x2+a2a)+c\int \sqrt{x^{2}+a^{2}} \: dx = \frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\ln \left ( \frac{x+\sqrt{x^{2}+a^{2}}}{a} \right )+c∫x2+a2​dx=2x​x2+a2​+2a2​ln(ax+x2+a2​​)+c, (xxiv) ∫x2−a2 dx=x2x2−a2−a22ln⁡(x+x2−a2a)+c\int \sqrt{x^{2}-a^{2}} \: dx = \frac{x}{2}\sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2}\ln \left ( \frac{x+\sqrt{x^{2}-a^{2}}}{a} \right )+c∫x2−a2​dx=2x​x2−a2​−2a2​ln(ax+x2−a2​​)+c, (xxv) ∫eax.sin⁡bx dx=eaxa2+b2(a sin⁡bx−bcos⁡bx)+c\int e^{ax}.\sin bx \: dx = \frac{e^{ax}}{a^{2}+b^{2}}\left ( a\, \sin bx-b\cos bx \right )+c∫eax.sinbxdx=a2+b2eax​(asinbx−bcosbx)+c, (xxvi) ∫eax.cos⁡bx dx=eaxa2+b2(a cos⁡bx+b sin⁡bx)+c\int e^{ax}.\cos bx \: dx = \frac{e^{ax}}{a^{2}+b^{2}}\left ( a\, \cos bx+b\: \sin bx \right )+c∫eax.cosbxdx=a2+b2eax​(acosbx+bsinbx)+c, (ii) ∫(f(x)±g(x)) dx=∫f(x) dx±g(x) dx\int (f(x)\pm g(x))\: dx = \int f(x)\: dx\pm g(x)\: dx∫(f(x)±g(x))dx=∫f(x)dx±g(x)dx, (iii) ∫ f(x) dx =  g(x)+c ⇒ ∫ f(ax+b) dx = g(ax+b)a+c\frac{g(ax+b)}{a}+cag(ax+b)​+c, If we substitute f(x) = t, then f’(x) dx = dt, ∫( f(x) g(x)) dx = f(x) ∫g(x) dx – ∫(ddx(f(x))∫(g(x))dx)dx\int\left ( \frac{d}{dx}(f(x))\int (g(x))dx\right ) dx∫(dxd​(f(x))∫(g(x))dx)dx.