72道积分题成就高手

1

$$\eqalign{ & \int {\frac{1}{{5x + 3}}{\kern 1pt} {\text{d}}x} \cr & {\text{u}} = {\text{5x}} + {\text{3}},{\text{d}}x = \frac{1}{5}{\kern 1pt} {\text{d}}u, \cr & = \frac{1}{5}\int {\frac{1}{u}{\kern 1pt} {\text{d}}u} \cr & = \frac{{\ln u}}{5} \cr & = \frac{{\ln \left| {5x + 3} \right|}}{5} + C \cr}$$
2

$$\eqalign{ & \int {{{\text{e}}^{2x + 3}}{\kern 1pt} {\text{d}}x} \cr & {\text{u}} = {\text{2x}} + {\text{3}},{\text{d}}x = \frac{1}{2}{\kern 1pt} {\text{d}}u, \cr & = \frac{1}{2}\int {{{\text{e}}^u}{\kern 1pt} {\text{d}}u} \cr & = \frac{{{{\text{e}}^{2x + 3}}}}{2} + C \cr} $$
3

$$\eqalign{ & \int {x{{\text{e}}^{{x^2}}}{\kern 1pt} {\text{d}}x} \cr & u = {x^2},{\text{d}}x = \frac{1}{{2x}}{\kern 1pt} {\text{d}}u \cr & = \frac{1}{2}\int {{{\text{e}}^u}{\kern 1pt} {\text{d}}u} \cr & = \frac{{{{\text{e}}^u}}}{2} \cr & = \frac{{{{\text{e}}^{{x^2}}}}}{2} + C \cr}$$
4

$$\eqalign{ & \int {x\sqrt {1 - {x^2}} {\kern 1pt} {\text{d}}x} \cr & u = 1 - {x^2},{\text{d}}x = - \frac{1}{{2x}}{\kern 1pt} {\text{d}}u, \cr & = - \frac{1}{2}\int {\sqrt u {\kern 1pt} {\text{d}}u} \cr & = - \frac{1}{2} \times \frac{{2{u^{\frac{3}{2}}}}}{3} \cr & = - \frac{{{u^{\frac{3}{2}}}}}{3} \cr & = - \frac{{{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}}}{3} + C \cr}$$
5

$$\eqalign{ & \int {\frac{{\sin \frac{1}{x}}}{{{x^2}}}{\kern 1pt} {\text{d}}x} \cr & u = \frac{1}{x},{\text{d}}x = - {x^2}{\kern 1pt} {\text{d}}u, \cr & = - \int {\sin u{\kern 1pt} {\text{d}}u} \cr & = \cos u \cr & = \cos \frac{1}{x} + C \cr} $$
6

$$\eqalign{ & \int {\frac{{{{\text{e}}^{3\sqrt x }}}}{{\sqrt x }}{\kern 1pt} {\text{d}}x} \cr & u = 3\sqrt x ,{\text{d}}x = \frac{{2\sqrt x }}{3}{\kern 1pt} {\text{d}}u, \cr & = \frac{2}{3}\int {{{\text{e}}^u}{\kern 1pt} {\text{d}}u} \cr & = \frac{{2{{\text{e}}^u}}}{3} \cr & = \frac{{2{{\text{e}}^{3\sqrt x }}}}{3} + C \cr}$$
7

$$\eqalign{ & \int {\frac{1}{{x\left( {1 + {x^6}} \right)}}{\kern 1pt} {\text{d}}x} \cr & = \int {\frac{1}{{\left( {\frac{1}{{{x^6}}} + 1} \right){x^7}}}{\kern 1pt} {\text{d}}x} \cr & u = \frac{1}{{{x^6}}} + 1,{\text{d}}x = - \frac{{{x^7}}}{6}{\kern 1pt} {\text{d}}u, \cr & = - \frac{1}{6}\int {\frac{1}{u}{\kern 1pt} {\text{d}}u} \cr & = - \frac{{\ln u}}{6} \cr & = - \frac{{\ln \left( {\frac{1}{{{x^6}}} + 1} \right)}}{6} + C \cr & = \ln \left| x \right| - \frac{{\ln \left( {1 + {x^6}} \right)}}{6} + C \cr} $$
8

$$\eqalign{ & \int {\cos \left( {2x} \right){\kern 1pt} {\text{d}}x} \cr & = \frac{1}{2}\int {\cos \left( {2x} \right){\kern 1pt} {\text{d}}\left( {2x} \right)} \cr & = \frac{{\sin \left( {2x} \right)}}{2} + C \cr}$$
9

$$\eqalign{ & \int {\frac{{\sin x}}{{\sqrt {5 + \cos x} }}{\kern 1pt} {\text{d}}x} \cr & u = 5 + \cos x,{\text{d}}x = - \frac{1}{{\sin x}}{\kern 1pt} {\text{d}}u, \cr & = - \int {\frac{1}{{\sqrt u }}{\kern 1pt} {\text{d}}u} \cr & = - 2\sqrt u \cr & = - 2\sqrt {5 + \cos x} + C \cr}$$
10

$$\eqalign{ & \int {{{\tan }^4}x{\kern 1pt} {\text{d}}x} \cr & = \int {{{\left( {{{\sec }^2}x - 1} \right)}^2}{\kern 1pt} {\text{d}}x} \cr & = \int {\left( {{{\sec }^4}x - 2{{\sec }^2}x + 1} \right){\text{d}}x} \cr & = \int {{{\sec }^4}x{\kern 1pt} {\text{d}}x} - 2\int {{{\sec }^2}x{\kern 1pt} {\text{d}}x} + \int {1{\kern 1pt} {\text{d}}x} \cr & = \int {{{\sec }^2}x\left( {{{\tan }^2}x + 1} \right){\kern 1pt} {\text{d}}x} - 2\tan x + x \cr & u = \tan x,{\text{d}}x = \frac{1}{{{{\sec }^2}x}}{\kern 1pt} {\text{d}}u, \cr & = \int {\left( {{u^2} + 1} \right){\text{d}}u} - 2\tan x + x \cr & = \frac{{{u^3}}}{3} + u - 2\tan x + x \cr & = \frac{{{{\tan }^3}x}}{3} - \tan x + x + C \cr}$$
11

$$\eqalign{ & \int {\frac{{{{\text{e}}^{2x}}}}{{{\text{1 + }}{{\text{e}}^x}}}{\kern 1pt} {\text{d}}x} \cr & u = {\text{1 + }}{{\text{e}}^x},{{\text{e}}^{2x}} = {\left( {u - 1} \right)^2}, \cr & {\text{d}}x = {{\text{e}}^{ - x}}{\kern 1pt} {\text{d}}u, \cr & = \int {\frac{{u - 1}}{u}{\kern 1pt} {\text{d}}u} \cr & = \int {1{\kern 1pt} {\text{d}}u} - \int {\frac{1}{u}{\kern 1pt} {\text{d}}u} \cr & = u - \ln u \cr & = {{\text{e}}^x} + 1 - \ln \left( {{\text{1 + }}{{\text{e}}^x}} \right) \cr & = {{\text{e}}^x} - \ln \left( {{\text{1 + }}{{\text{e}}^x}} \right) + C \cr} $$
12

$$\eqalign{ & \int {\frac{1}{{{\text{1 + }}{{\text{e}}^x}}}{\kern 1pt} {\text{d}}x} \cr & u = {\text{1 + }}{{\text{e}}^x},{\text{d}}x = {{\text{e}}^{ - x}}{\kern 1pt} {\text{d}}u, \cr & = \int {\frac{1}{{\left( {u - 1} \right)u}}{\kern 1pt} {\text{d}}u} \cr & \frac{1}{{\left( {u - 1} \right)u}} = \frac{A}{u} + \frac{B}{{u - 1}}, \cr & A = - 1,B = 1, \cr & = \int {\left( {\frac{1}{{u - 1}} - \frac{1}{u}} \right){\text{d}}u} \cr & = \ln \left( {u - 1} \right) - \ln u \cr & = x - \ln \left( {{\text{1 + }}{{\text{e}}^x}} \right) + C \cr} $$
13

$$\eqalign{ & \int {\frac{1}{{x{{\ln }^2}x}}{\kern 1pt} {\text{d}}x} \cr & u = \ln x,{\text{d}}x = x{\kern 1pt} {\text{d}}u, \cr & = \int {\frac{1}{{{u^2}}}{\kern 1pt} {\text{d}}u} \cr & = - \frac{1}{u} \cr & = - \frac{1}{{\ln x}} + C \cr}$$
14

$$\eqalign{ & \int {\frac{1}{{x\left( {1 + 2\ln x} \right)}}{\kern 1pt} {\text{d}}x} \cr & u = 1 + 2\ln x,{\text{d}}x = \frac{x}{2}{\kern 1pt} {\text{d}}u, \cr & = \frac{1}{2}\int {\frac{1}{u}{\kern 1pt} {\text{d}}u} \cr & = \frac{{\ln u}}{2} \cr & = \frac{{\ln \left| {1 + 2\ln x} \right|}}{2} + C \cr}$$
15

$$\eqalign{ & \int {\frac{1}{{{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}{\kern 1pt} {\text{d}}x} \cr & = \int {{{\sec }^2}x\frac{1}{{{b^2}{{\tan }^2}x + {a^2}}}{\kern 1pt} {\text{d}}x} \cr & u = \tan x,{\text{d}}x = \frac{1}{{{{\sec }^2}x}}{\kern 1pt} {\text{d}}u, \cr & = \int {\frac{1}{{{b^2}{u^2} + {a^2}}}{\kern 1pt} {\text{d}}u} \cr & v = \frac{{bu}}{a},{\text{d}}u = \frac{a}{b}{\kern 1pt} {\text{d}}v, \cr & = \frac{1}{{ab}}\int {\frac{1}{{{v^2} + 1}}{\kern 1pt} {\text{d}}v} \cr & = \frac{{\arctan v}}{{ab}} \cr & = \frac{{\arctan \left( {\frac{{bu}}{a}} \right)}}{{ab}} \cr & = \frac{{\arctan \left( {\frac{{b\tan x}}{a}} \right)}}{{ab}} + C \cr} $$
16

$$\eqalign{ & \int {\frac{1}{{{x^2} + {a^2}}}{\kern 1pt} {\text{d}}x} \cr & u = \frac{x}{a}, \cr & = \frac{1}{a}\int {\frac{1}{{{u^2} + 1}}{\kern 1pt} {\text{d}}u} \cr & = \frac{{\arctan u}}{a} \cr & = \frac{{\arctan \frac{x}{a}}}{a} + C \cr}$$
17

$$\eqalign{ & \int {\frac{1}{{{a^2} - {x^2}}}{\kern 1pt} {\text{d}}x} \cr & = - \int {\frac{1}{{{x^2} - {a^2}}}{\kern 1pt} {\text{d}}x} \cr & = - \int {\frac{1}{{\left( {x - a} \right)\left( {x + a} \right)}}{\kern 1pt} {\text{d}}x} \cr & \frac{1}{{\left( {x - a} \right)\left( {x + a} \right)}} = \frac{A}{{x + a}} + \frac{B}{{x - a}} \cr & Ax - Aa + Bx + Ba = 1 \cr & A + B = 0,Ba - Aa = 1, \cr & A = - \frac{1}{{2a}},B = \frac{1}{{2a}}, \cr & = - \int {\left[ {\frac{1}{{2a\left( {x - a} \right)}} - \frac{1}{{2a\left( {x + a} \right)}}} \right]{\text{d}}x} \cr & = - \frac{1}{{2a}}\int {\frac{1}{{x - a}}{\kern 1pt} {\text{d}}x} + \frac{1}{{2a}}\int {\frac{1}{{x + a}}{\kern 1pt} {\text{d}}x} \cr & = \frac{{\ln \left| {x + a} \right|}}{{2a}} - \frac{{\ln \left| {x - a} \right|}}{{2a}} + C \cr & = \frac{{\ln \left| {x + a} \right| - \ln \left| {x - a} \right|}}{{2a}} + C \cr}$$
18

$$\eqalign{ & \int {\frac{1}{{\sqrt {{a^2} - {x^2}} }}{\kern 1pt} {\text{d}}x} \cr & u = \frac{x}{a},{\text{d}}x = a{\kern 1pt} {\text{d}}u, \cr & = \int {\frac{1}{{\sqrt {1 - {u^2}} }}{\kern 1pt} {\text{d}}u} \cr & = \arcsin u \cr & = \arcsin \frac{x}{a} + C \cr} $$
19

$$\eqalign{ & \int {{{\sin }^3}x{\kern 1pt} {\text{d}}x} \cr & = \int {\left( {1 - {{\cos }^2}x} \right)\sin x{\kern 1pt} {\text{d}}x} \cr & u = \cos x,{\text{d}}x = - \frac{1}{{\sin x}}{\kern 1pt} {\text{d}}u, \cr & = \int {\left( {{u^2} - 1} \right){\text{d}}u} \cr & = \frac{{{u^3}}}{3} - u \cr & = \frac{{{{\cos }^3}x}}{3} - \cos x + C \cr}$$
20

$$\eqalign{ & \int {{{\sin }^5}x{\kern 1pt} {\text{d}}x} \cr & = \int {{{\left( {1 - {{\cos }^2}x} \right)}^2}\sin x{\kern 1pt} {\text{d}}x} \cr & u = \cos x,{\text{d}}x = - \frac{1}{{\sin x}}{\kern 1pt} {\text{d}}u, \cr & = \int { - {{\left( {1 - {u^2}} \right)}^2}{\kern 1pt} {\text{d}}u} \cr & = - \int {\left( {{u^4} - 2{u^2} + 1} \right){\text{d}}u} \cr & = - \left( {\frac{{{u^5}}}{5} - \frac{{2{u^3}}}{3} + u} \right) \cr & = - \frac{{{u^5}}}{5} + \frac{{2{u^3}}}{3} - u \cr & = - \frac{{{{\cos }^5}x}}{5} + \frac{{2{{\cos }^3}x}}{3} - \cos x + C \cr & \cr}$$
21

$$\eqalign{ & \int {{{\cos }^3}x{\kern 1pt} {\text{d}}x} \cr & = \int {\cos x\left( {1 - {{\sin }^2}x} \right){\kern 1pt} {\text{d}}x} \cr & u = \sin x,{\text{d}}x = \frac{1}{{\cos x}}{\kern 1pt} {\text{d}}u, \cr & = \int {\left( {1 - {u^2}} \right){\text{d}}u} \cr & = u - \frac{{{u^3}}}{3} \cr & = \sin x - \frac{{{{\sin }^3}x}}{3} + C \cr}$$