Một số phương pháp tìm nguyên hàm (buổi 2)

Đặt: \(\left\{ \begin{array}{l} u = x\\ dv = {e^x}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = dx\\ v = {e^x} \end{array} \right.\)
Vậy: \(\int {x.{e^x}dx} = x{e^x} - \int {{e^x}dx} = x{e^x} - {e^x} + C.\)

 

Đặt: \(u = \sin x + 1 \Rightarrow du = \cos xdx\)
Vậy: \(\int {{{(\sin x + 1)}^3}\cos x} dx = \int {{u^3}du} = \frac{1}{4}{u^4} + C = \frac{1}{4}{(\sin x + 1)^4} + C.\)

 

Đặt \(\left\{ \begin{array}{l} u = x\\ dv = \cos 2xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = dx\\ v = \frac{1}{2}\sin 2x \end{array} \right.\)
Vậy:
\(\begin{array}{l} \int {x\cos 2xdx} = \frac{1}{2}x.\sin 2x - \frac{1}{2}\int {\sin 2xdx} \\ = \frac{1}{2}x.\sin 2x - \frac{1}{4}\cos 2x + C. \end{array}\)

 

Ta có: \(\int {f(x)dx} = \int {\frac{{2x + 1}}{{{x^2} + x + 1}}dx}\)
Đặt: \(u = {x^2} + x + 1 \Rightarrow du = \left( {2x + 1} \right)dx\)
Vậy: \(\int {f(x)dx} = \int {\frac{1}{u}du} = \ln \left| u \right| + C = \ln \left| {{x^2} + x + 1} \right| + C = \ln ({x^2} + x + 1) + C\)
Ta có: \(F(2) = 3 \Rightarrow \ln 7 + C = 3 \Rightarrow C = 3 - \ln 7\)
Do đó: \(F\left( 1 \right) = \ln 3 + 3 - \ln 7 = 3 - \ln \frac{7}{3}.\)

 

Ta có \(\int {f(x)dx} = \int {\tan xdx} = \int {\frac{{\sin xdx}}{{\cos x}}}\)
Đặt \(u = \cos \Rightarrow du = - \sin xdx\)
Vậy \(\int {f(x)dx} = - \int {\frac{1}{u}du = - \ln \left| u \right|} = - \ln \left| {\cos x} \right| + C\)

 

Đặt: \(u = 3x + 1 \Rightarrow du = 3dx\)
Vậy: \(\int {\cos \left( {3x + 1} \right)dx} = \frac{1}{3}\int {\cos udu} = \frac{1}{3}\sin u + C = \frac{1}{3}\sin \left( {3x + 1} \right) + C.\)

 

Đặt: \(u = \sqrt[3]{{x + 1}} \Rightarrow {u^3} = x + 1 \Rightarrow 3{u^2}du = dx\)
Vậy: \(\int {f\left( x \right)dx} = \int {\sqrt[3]{{x + 1}}dx} = 3\int {u.{u^2}du = \frac{3}{4}{u^4} + C} = \frac{3}{4}{\left( {x + 1} \right)^{\frac{4}{3}}} + C.\)

 

Ta có \(F\left( x \right) = \int {f\left( x \right)dx = \int {{4^x}{{.2}^{2x + 3}}dx = } \int {{{4.2}^{2x}}{{.2}^{2x + 1}}dx} } = \int 4 {.2^{4x + 1}}dx\)
Đặt: \(u = 4x + 1 \Rightarrow du = 4dx\)
Vậy: \(\int {f(x)dx} = \int {{2^u}du} = \frac{{{2^u}}}{{\ln 2}} + C = \frac{{{2^{4x + 1}}}}{{\ln 2}} + C.\)

 

\(\int {f(x)dx} = \int {\frac{x}{{{x^2} + 1}}dx}\)
Đặt: \(u = {x^2} + 1 \Rightarrow du = 2xdx\)
Vậy: \(\int {f(x)dx} = \frac{1}{2}\int {\frac{1}{u}du} = \frac{1}{2}\ln \left| u \right| + C = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C\)
Do F(0) = 1 nên C=1
Vậy: \(F\left( 1 \right) = \frac{1}{2}\ln 2 + 1.\)

 

\(\int {f(x)dx} = \int {{3^x}\sqrt {{3^x} + 1} dx}\)
Đặt \(u = \sqrt {{3^x} + 1} \Rightarrow {u^2} = 3x + 1 \Rightarrow 2udu = {3^x}\ln 3dx \Rightarrow \frac{2}{{\ln 3}}udu = {3^x}dx\) khi đó:
\(\begin{array}{l} \int {f(x)dx} = \frac{2}{{\ln 3}}\int {u.udu} = \frac{2}{{\ln 3}}\int {{u^2}du} = \frac{2}{{3\ln 3}}{u^3} + C\\ = \frac{2}{{3\ln 3}}\sqrt {{{\left( {{3^x} + 1} \right)}^3}} + C = \frac{2}{{3\ln 3}}({3^x} + 1)\sqrt {\left( {{3^x} + 1} \right)} + C. \end{array}\)

 

\(\int\limits_0^{\frac{\pi }{2}} {f\left( x \right)} = \int\limits_0^{\frac{\pi }{2}} {\frac{{2\sin 2x\cos 2x}}{{1 + \frac{{1 + \cos 2x}}{2}}}dx} = 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\cos 2x.\sin 2x}}{{3 + \cos 2x}}dx}\)
Đặt: \(t = \cos 2x \Rightarrow dt = - 2\sin 2x\)
\(\Rightarrow I = - 2\int\limits_1^{ - 1} {\frac{t}{{t + 3}}dx} = 2\int\limits_{ - 1}^1 {\frac{{t + 3 - 3}}{{t + 3}}dt} = 2\int\limits_{ - 1}^1 {\left( {1 - \frac{3}{{t + 3}}} \right)dt}\)
\(= \left. {\left( {2t - 6\ln \left| {t + 3} \right|} \right)} \right|_{ - 1}^1 = 4 - 6\ln 2.\)
\(F\left( {\frac{\pi }{2}} \right) - F\left( 0 \right) = 4 - 6\ln 2 \Rightarrow F\left( 0 \right) = - 4 + 6\ln 2.\)

 

Xét nguyên hàm: \(\int {2x{{({x^2} + 1)}^4}dx}\)
Đặt: \(u = {x^2} + 1 \Rightarrow du = 2xdx\)
Khi đó: \(\int {2x{{({x^2} + 1)}^4}dx} = \int {{u^4}du} = \frac{1}{5}{u^5} + C = \frac{1}{5}{({x^2} + 1)^5} + C\)
Khi đó \(F\left( 1 \right) = \frac{{32}}{5} + C = 6 \Rightarrow C = - \frac{2}{5}\).
Vậy \(F\left( x \right) = \frac{{{{({x^2} + 1)}^5}}}{5} - \frac{2}{5}\).

 

Xét \(\int {f\left( x \right)} .{\rm{d}}x = \int {\sqrt {l{n^2}x + 1} .\frac{{lnx}}{x}} .{\rm{d}}x\).
Đặt \(\sqrt {l{n^2}x + 1} = t\)
\(\Rightarrow l{n^2}x = {t^2} - 1 \Rightarrow \frac{{lnx}}{x}.{\rm{d}}x = t.{\rm{d}}t\)
Vì vậy: \(F\left( x \right) = \int {{t^2}d} = \frac{1}{3}{t^3} + C = \frac{{\sqrt {{{\left( {l{n^2}x + 1} \right)}^3}} }}{3} + C.\)
Do \(F\left( 1 \right) = \frac{1}{3} \Rightarrow C = 0\). Vậy \({\left[ {F\left( e \right)} \right]^2} = \frac{8}{9}.\)

 

Xét nguyên hàm: \(\int {\frac{1}{{{x^2}}}\cos \frac{2}{x}dx}\)
Đặt \(t = \frac{2}{x} \Rightarrow dt = - \frac{2}{{{x^2}}}dx\)
Vậy: \(\int {\frac{1}{{{x^2}}}\cos \frac{2}{x}dx} = - \frac{1}{2}\int {\cos tdt} = - \frac{1}{2}\sin t + C = - \frac{1}{2}\sin \frac{2}{x} + C.\)

 

Xét nguyên hàm: \(\int {\sqrt[3]{{3x + 1}}dx}\)
Đặt \(t = \sqrt[3]{{3x + 1}} \Rightarrow {t^3} = 3x + 1 \Rightarrow 3{t^2} = 3dx\)
Suy ra:
\(\begin{array}{l} \int {\sqrt[3]{{3x + 1}}dx} = \int {{t^2}.tdt} = \frac{1}{4}{t^4} + C\\ = \frac{1}{4}\sqrt[3]{{{{(3x + 1)}^4}}} + C = \frac{1}{4}(3x + 1)\sqrt[3]{{3x + 1}} + C. \end{array}\)

 

Ta có: \(\int {f\left( x \right)d{\rm{x}}} = \int {{{\left( {2{\rm{x}} + 1} \right)}^9}d{\rm{x}}} \)
Đặt: \(t = 2x + 1 \Rightarrow dt = 2dx\)
Ta có: \(\int {f\left( x \right)d{\rm{x}}} = \int {{{\left( {2{\rm{x}} + 1} \right)}^9}d{\rm{x}}} = \frac{1}{2}\int {{t^9}dt} = \frac{1}{{20}}{t^{10}} + C = \frac{1}{{20}}{\left( {2{\rm{x}} + 1} \right)^{10}} + C.\)

 

Xét nguyên hàm: \(\int {f(x)dx} = \int {\frac{1}{{x\ln x}}dx} \)
Đặt \(t = \ln x \Rightarrow dt = \frac{1}{x}dx\)
Khi đó: \(\int {f(x)dx} = \int {\frac{1}{t}dt} = \ln \left| t \right| + C = \ln \left| {\ln x} \right| + C\)
Ta có: \(F\left( e \right) = 3 = \ln \left| {\ln e} \right| + C \Rightarrow C = 2\)
Vậy: \(F\left( {\frac{1}{e}} \right) = \ln \left| {\ln \left( {\frac{1}{e}} \right)} \right| + 2 = 3.\)

 

Xét nguyên hàm \(\int {3x.{e^{{x^2}}}} dx\)
Đặt \(t = {x^2} \Rightarrow dt = 2xdx \Rightarrow \frac{3}{2}dt = 3xdx\)
Vậy: \(\int {3x.{e^{{x^2}}}} dx = \int {\frac{3}{2}{e^t}dt} = \frac{3}{2}{e^t} + C = \frac{3}{2}{e^{{x^2}}} + C\)

 

Ta có \(I = \int y dx = \int {\frac{{{x^2} + 1}}{{{x^3} - x}}} dx = \int {\frac{{1 + \frac{1}{{{x^2}}}}}{{x - \frac{1}{x}}}} dx\)
Đặt: \(t = x - \frac{1}{{{x^2}}} \Rightarrow dt = \left( {1 + \frac{1}{{{x^2}}}} \right)dx\)
Suy ra: \(I = \int {\frac{1}{t}dt} = \ln \left| t \right| + C = \ln \left| {x - \frac{1}{x}} \right| + C = \ln \left| {\frac{{{x^2} - 1}}{x}} \right| + C.\)
Với \(C = \ln 2\)
Ta có \(I = \ln \left| {\frac{{{x^2} - 1}}{x}} \right| + \ln 2 = \ln \left| {\frac{{2{x^2} - 2}}{x}} \right|.\)
Với \(C = - \ln 2\)
Ta có \(I = \ln \left| {\frac{{{x^2} - 1}}{x}} \right| - \ln 2 = \ln \left| {\frac{{{x^2} - 1}}{{2x}}} \right|.\)
Với C=0
Ta có \(I = \ln \left| {\frac{{{x^2} - 1}}{x}} \right|.\)

 

\(I = \int {f\left( x \right)dx} = \int {\frac{1}{{{{\sin }^2}2x}}} dx \).
Đặt: \(t = 2x \Rightarrow dt = 2dx\)
Suy ra: \(I = \frac{1}{2}\int {\frac{1}{{{{\sin }^2}t}}dt} = - \frac{1}{2}\cot t + C = - \frac{1}{2}\cot 2x + C\)