Nguyên hàm

Trên khoảng $\left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right)$ ta có:

 $\begin{array}{l}f\left( x \right) = \int\limits_{}^{} {f'\left( x \right)dx}  = \int\limits_{}^{} {\frac{1}{{{x^2} - 1}}dx}  = \frac{1}{2}\int\limits_{}^{} {\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right)dx}  = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + {C_1} = \frac{1}{2}\ln \frac{{x - 1}}{{x + 1}} + {C_1}\\f\left( { - 3} \right) + f\left( 3 \right) = \frac{1}{2}\ln 2 + \frac{1}{2}\ln \frac{1}{2} + {C_1} = 0 \Leftrightarrow {C_1} = 0\end{array}$

Trên khoảng $\left( { - 1;1} \right)$ ta có:

$\begin{array}{l}f\left( x \right) = \int\limits_{}^{} {f'\left( x \right)dx}  = \int\limits_{}^{} {\frac{1}{{{x^2} - 1}}dx}  = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + {C_2} = \frac{1}{2}\ln \frac{{ - x + 1}}{{x + 1}} + {C_2}\\f\left( { - \frac{1}{2}} \right) + f\left( {\frac{1}{2}} \right) = \frac{1}{2}\ln 3 + \frac{1}{2}\ln \frac{1}{3} + {C_2} = 2 \Leftrightarrow {C_2} = 2\\ \Rightarrow f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{2}\ln \frac{{x - 1}}{{x + 1}}\,\,khi\,\,x \in \left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right)\\\frac{1}{2}\ln \frac{{ - x + 1}}{{x + 1}} + 2\,\,khi\,\,x \in \left( { - 1;1} \right)\end{array} \right.\\\Rightarrow T = f\left( { - 2} \right) + f\left( 0 \right) + f\left( 5 \right) = \frac{1}{2}\ln 3 + \frac{1}{2}\ln 1 + 2 + \frac{1}{2}\ln \frac{2}{3} = \ln \sqrt 2  + 2\end{array}$

Ta có $f\left( x \right) = \left| {1 + x} \right| - \left| {1 - x} \right| = \left\{ \begin{array}{l} 2\,\,\,\,\,\,\,\,\,\,khi\,\,\,\,x \ge 1\\ 2x\,\,\,\,\,\,\,\,khi\,\,\,\, - \,1 \le x < 1\\ - \,2\,\,\,\,\,\,\,khi\,\,\,\,\,x < - \,1 \end{array} \right. \Rightarrow F\left( x \right) = \left\{ \begin{array}{l} 2x + {C_1}\,\,\,\,\,\,\,\,khi\,\,\,\,x \ge 1\\ {x^2} + {C_2}\,\,\,\,\,\,\,\,khi\,\,\,\, - \,1 \le x < 1\\ - \,2x + {C_3}\,\,\,\,khi\,\,\,\,\,x < - \,1 \end{array} \right.$

Theo đề bài ta có $\left\{ \begin{array}{l} F\left( 1 \right) = 3\\ F\left( { - 1} \right) = 2\\ F\left( { - 2} \right) = 4 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2 + {C_1} = 3\\ 1 + {C_2} = 2\\ 4 + {C_3} = 4 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {C_1} = 1\\ {C_2} = 1\\ {C_3} = 0 \end{array} \right.$

$\begin{array}{l} \Rightarrow F\left( x \right) = \left\{ \begin{array}{l} 2x + 1\,\,\,\,\,\,\,\,khi\,\,\,\,x \ge 1\\ {x^2} + 1\,\,\,\,\,\,\,\,khi\,\,\,\, - \,1 \le x < 1\\ - \,2x\,\,\,\,khi\,\,\,\,\,x < - \,1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} F\left( 2 \right) = 2.2 + 1 = 5\\ F\left( 0 \right) = 1\\ F\left( { - 3} \right) = - 2.\left( { - 3} \right) = 6. \end{array} \right.\\ \Rightarrow T = 5 + 1 + 6 = 12. \end{array}$

$\begin{array}{l}f'(x) = \frac{1}{{{x^2} - 1}} \Rightarrow \int {f'(x)dx}  = \int {\frac{1}{{{x^2} - 1}}dx}  \Rightarrow f(x) = \int {\frac{1}{{(x - 1)(x + 1)}}dx}  = \frac{1}{2}\int {\frac{1}{{x - 1}}dx + } \frac{1}{2}\int {\frac{1}{{x + 1}}dx = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C} \\ \Rightarrow f(x) = \left\{ \begin{array}{l}\frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + {C_1},\left[ \begin{array}{l}x <  - 1\\x > 1\end{array} \right.\\\frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + {C_2}, - 1 < x < 1\end{array} \right.\end{array}$

Ta có:

$f(-3)+f(3)=0\Leftrightarrow \frac{1}{2}\ln 2+{{C}_{1}}+\frac{1}{2}\ln \frac{1}{2}+{{C}_{1}}=0\Leftrightarrow 2{{C}_{1}}=0\Leftrightarrow {{C}_{1}}=0$

$f\left( -\frac{1}{2} \right)+f\left( \frac{1}{2} \right)=2\Leftrightarrow \frac{1}{2}\ln 3+{{C}_{2}}+\frac{1}{2}\ln \frac{1}{3}+{{C}_{2}}=2\Leftrightarrow {{C}_{2}}=1$

$P=f(0)+f(4)=\left( \frac{1}{2}\ln \left| \frac{0-1}{0+1} \right|+1 \right)+\left( \frac{1}{2}\ln \left| \frac{4-1}{4+1} \right| \right)=\frac{1}{2}\ln \frac{3}{5}+1$

Ta có $f\left( x \right)=\int{{f}'\left( x \right)\,\text{d}x}=\int{\frac{\text{d}x}{{{x}^{2}}-1}}=\frac{1}{2}\ln \left| \frac{x-1}{x+1} \right|+C=\left\{ \begin{align} & \frac{1}{2}\ln \frac{x-1}{x+1}+{{C}_{1}}\,\,\,\,\,khi\,\,\,x>1 \\ & \frac{1}{2}\ln \frac{1-x}{x+1}+{{C}_{2}}\,\,\,\,\,khi\,\,\,-\,1<x<1 \\ & \frac{1}{2}\ln \frac{x-1}{x+1}+{{C}_{3}}\,\,\,\,\,khi\,\,\,x<-\,1 \\ \end{align} \right..$

Suy ra $f\left( -\,3 \right)+f\left( 3 \right)=0\Leftrightarrow \,\,\frac{1}{2}\ln 2+{{C}_{1}}+\frac{1}{2}\ln \frac{1}{2}+{{C}_{3}}=0\Leftrightarrow \,\,{{C}_{1}}+{{C}_{3}}=0.$ Và $f\left( -\frac{1}{2} \right)+f\left( \frac{1}{2} \right)=2\Leftrightarrow \,\,\frac{1}{2}\ln 3+{{C}_{2}}+\frac{1}{2}\ln \frac{1}{3}+{{C}_{2}}=2\Leftrightarrow \,\,{{C}_{2}}=1.$

Vậy $T=f\left( -\,2 \right)+f\left( 0 \right)+f\left( 5 \right)=\frac{1}{2}\ln 3+{{C}_{3}}+{{C}_{2}}+\frac{1}{2}\ln \frac{2}{3}+{{C}_{1}}=\frac{1}{2}\ln 2+1.$

$F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)$ nên ta có $F'\left( x \right) = f\left( x \right)$

Ta có:

$\begin{array}{l}F'\left( x \right) = \left( {3a{x^2} + 2bx + c} \right){e^x} + \left( {a{x^3} + b{x^2} + cx + d} \right){e^x}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {a{x^3} + \left( {3a + b} \right){x^2} + \left( {2b + c} \right)x + c + d} \right){e^x}\end{array}$

Do đó $\left( {a{x^3} + \left( {3a + b} \right){x^2} + \left( {2b + c} \right)x + c + d} \right){e^x} = \left( {2{x^3} + 9{x^2} - 2x + 5} \right){e^x}$

Đồng nhất hệ số ta có: $\left\{ \begin{array}{l}a = 2\\3a + b = 9\\2b + c =  - 2\\c + d = 5\end{array} \right.\left\{ \begin{array}{l}a = 2\\b = 3\\c =  - 8\\d = 13\end{array} \right. \Rightarrow {a^2} + {b^2} + {c^2} + {d^2} = 246$

Ta có  $\left[ f\left( x \right).f'\left( x \right) \right]'={{\left[ f'\left( x \right) \right]}^{2}}+f\left( x \right).f''\left( x \right)=15{{x}^{4}}+12x$

Nguyên hàm 2 vế ta được $f\left( x \right).f'\left( x \right)=3{{x}^{5}}+6{{x}^{2}}+C$

Do $f\left( 0 \right)=f'\left( 0 \right)=1\Rightarrow C=1$

Tiếp tục nguyên hàm 2 vế ta được: $\int{f\left( x \right)df\left( x \right)}=\int{\left( 3{{x}^{5}}+6{{x}^{2}}+1 \right)dx}$

$\Rightarrow \frac{{{f}^{2}}\left( x \right)}{2}=\frac{3{{x}^{6}}}{6}+\frac{6{{x}^{3}}}{3}+x+D=\frac{1}{2}{{x}^{6}}+2{{x}^{3}}+x+D$

Do $f\left( 0 \right)=1\Rightarrow D=\frac{1}{2}\Rightarrow \frac{{{f}^{2}}\left( x \right)}{2}=\frac{1}{2}{{x}^{6}}+2{{x}^{3}}+x+\frac{1}{2}\Rightarrow {{f}^{2}}\left( 1 \right)=8$

$f'(x)\sqrt{{{x}^{2}}+1}=2x\sqrt{f(x)+1}\Leftrightarrow \frac{f'(x)}{\sqrt{f(x)+1}}=\frac{2x}{\sqrt{{{x}^{2}}+1}}\Rightarrow \int{\frac{f'(x)}{\sqrt{f(x)+1}}}dx=\int{\frac{2x}{\sqrt{{{x}^{2}}+1}}}dx\Leftrightarrow \int{\frac{d\left( f(x)+1 \right)}{\sqrt{f(x)+1}}}=\int{\frac{d({{x}^{2}}+1)}{\sqrt{{{x}^{2}}+1}}}$

$\Leftrightarrow 2\sqrt{f(x)+1}=2\sqrt{{{x}^{2}}+1}+C$

Mà $f(0)=0\Rightarrow 2\sqrt{0+1}=2\sqrt{{{0}^{2}}+1}+C\Rightarrow C=0$

$\Rightarrow \sqrt{f(x)+1}=\sqrt{{{x}^{2}}+1}\Leftrightarrow f(x)={{x}^{2}}$

$\Rightarrow f\left( \sqrt{3} \right)={{\left( \sqrt{3} \right)}^{2}}=3$

$f\left( x \right) = f'\left( x \right)\sqrt {3x + 1}  \Leftrightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \frac{1}{{\sqrt {3x + 1} }}$

Lấy nguyên hàm hai vế ta có $\int\limits_{}^{} {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx}  = \int\limits_{}^{} {\frac{{dx}}{{\sqrt {3x + 1} }} \Leftrightarrow \ln \left| {f\left( x \right)} \right|}  = \frac{2}{3}\sqrt {3x + 1}  + C = \ln f\left( x \right)\,\,\,\left( {f\left( x \right) > 0} \right)$

$\begin{array}{l} \Rightarrow \ln \left| {f\left( 1 \right)} \right| = \frac{2}{3}.\sqrt 4  + C \Leftrightarrow C =  - \frac{4}{3} \Rightarrow \ln f\left( x \right) = \frac{2}{3}\sqrt {3x + 1}  - \frac{4}{3} \Leftrightarrow f\left( x \right) = {e^{\frac{2}{3}\sqrt {3x + 1}  - \frac{4}{3}}}\\ \Rightarrow f\left( 5 \right) = {e^{\frac{4}{3}}} \approx 3,79\end{array}$

Chuyển vế và nhân cả hai vế với ${e^{ - x}}$ ta có:

$\begin{array}{l}f'\left( x \right) = f\left( x \right) + {x^2}{e^x} + 1\,\,\forall x \in R\\ \Leftrightarrow f'\left( x \right){e^{ - x}} - {e^{ - x}}f\left( x \right) = {x^2} + {e^{ - x}}\end{array}$

Ta có $\left[ {f\left( x \right){e^{ - x}}} \right]' = f'\left( x \right){e^{ - x}} - {e^{ - x}}f\left( x \right)$

$\Rightarrow \left[ {f\left( x \right){e^{ - x}}} \right]' = {x^2} + {e^{ - x}}$

Lấy nguyên hàm hai vế ta được $f\left( x \right){e^{ - x}} = \frac{{{x^3}}}{3} - {e^{ - x}} + C \Rightarrow f\left( x \right) = \frac{{{x^3}{e^x}}}{3} - 1 + C{e^x}$

Ta có $f\left( 0 \right) =  - 1 \Leftrightarrow  - 1 + C =  - 1 \Leftrightarrow C = 0 \Rightarrow f\left( x \right) = \frac{{{x^3}{e^x}}}{3} - 1$

$\Rightarrow f\left( 3 \right) = \frac{{{3^3}.{e^3}}}{3} - 1 = 9{e^3} - 1$

Ta có: $\int {\left( {{e^x} + 1} \right)dx = {e^x} + x + C}$

$\int {f\left( x \right)dx}  = \int {\left( {3{x^2} + \sin x} \right)dx}  = {x^3} - \cos x + C$

$\int {\left( {3{x^2} + \dfrac{1}{{{{\cos }^2}x}}} \right)} dx = {x^3} + \tan x + C$.

Ta có: $\int {{x^5}dx}  = \dfrac{{{x^6}}}{6} + C.$

Ta có $f\left( x \right) = \int {f'\left( x \right) = \int {\frac{2}{{2x - 1}}dx} }$

$\Rightarrow f\left( x \right) = \ln \left| {2x - 1} \right| + C$

+) $f\left( x \right) = \ln \left( {2x - 1} \right) + C$$\left( {x \ge \frac{1}{2}} \right)$

Mà $f\left( 1 \right) = 2 \Rightarrow f\left( x \right) = \ln \left( {2x - 1} \right) + 2$

$+ )f\left( x \right) = \ln \left( {1 - 2x} \right) + C$$\left( {x < \frac{1}{2}} \right)$

Mà $f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = \ln \left( {1 - 2x} \right) + 1$

Với $x =  - 1 \Rightarrow f\left( { - 1} \right) = \ln \left( {1 - 2x} \right) + 1 = \ln 3 + 1$

Với $x = 3 \Rightarrow f\left( 3 \right) = \ln \left( {2x - 1} \right) + 2 = \ln 5 + 2$

$\Rightarrow f\left( { - 1} \right) + f\left( 3 \right) = \ln 15 + 3$

Lấy nguyên hàm hai vế biểu thức $\dfrac{{f'\left( x \right)}}{{f\left( x \right)}} = 2 - 2x$ ta có:

$\begin{array}{l}\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx}  = \int {\left( {2 - 2x} \right)dx} \\ \Leftrightarrow \ln \left| {f\left( x \right)} \right| =  - {x^2} + 2x + C\end{array}$

Theo bài ra ta có $f\left( 0 \right) = 1$ $\Leftrightarrow \ln \left| {f\left( 0 \right)} \right| = C \Leftrightarrow \ln 1 = C \Leftrightarrow C = 0$.

$\begin{array}{l} \Leftrightarrow \ln \left| {f\left( x \right)} \right| =  - {x^2} + 2x\\ \Leftrightarrow \left| {f\left( x \right)} \right| = {e^{ - {x^2} + 2x}}\\ \Leftrightarrow f\left( x \right) = {e^{ - {x^2} + 2x}}\,\,\left( {Do\,\,f\left( x \right) > 0\,\,\forall x \in \mathbb{R}} \right)\end{array}$

Ta có : $f'\left( x \right) = \left( { - 2x + 2} \right){e^{ - {x^2} + 2x}} = 0 \Leftrightarrow x = 1$.

BBT:

Phương trình $f\left( x \right) = m$ có hai nghiệm thực phân biệt khi và chỉ khi đường thẳng $y = m$ cắt đồ thị hàm số $y = f\left( x \right)$ tại hai điểm phân biệt, dựa vào BBT ta suy ra $0 < m < e$.

Ta có $\int {f\left( x \right)dx}  = \int {\cos 5xdx}  = \dfrac{1}{5}\sin 5x + C.$

$\int {f\left( x \right)dx}  = \int {\sin 2xdx}  =  - \dfrac{1}{2}\cos 2x + C$.

Ta có: $F\left( x \right) = \int {\sin 2xdx}  =  - \frac{1}{2}\cos 2x + C$ $\Rightarrow$ đáp án C đúng.

Lại có: $- \frac{1}{2}\cos 2x + C =  - \frac{1}{2}\left( {2{{\cos }^2}x - 1} \right) + C$$=  - {\cos ^2}x + C'$$\Rightarrow$ đáp án A đúng.

$- \frac{1}{2}\cos 2x + C =  - \frac{1}{2}\left( {1 - 2{{\sin }^2}x} \right) + C$$= {\sin ^2}x + C'$ $\Rightarrow$ đáp án B đúng.

Ta có

$\begin{array}{l}f\left( x \right) = 3{x^2} + 8\sin x\\ \Rightarrow \int {f\left( x \right)dx = \int {3{x^2}dx + \int {8\sin xdx} } } \\ \Rightarrow \int {f\left( x \right)dx}  = {x^3} - 8\cos x + C.\end{array}$

Ta có: $\int {\dfrac{{dx}}{x}}  = \ln \left| x \right| + C = \ln x + C\,\,\left( {do\,\,x > 0} \right)$.

Dựa vào các đáp án ta thấy:

Đáp án A: $\dfrac{1}{2}\ln {x^2} = \ln \left| x \right| = \ln x$ là 1 nguyên hàm của hàm số $f\left( x \right) = \dfrac{1}{x}$ khi $C = 0$.

Đáp án B: $\ln x$ là một nguyên hàm của hàm số $f\left( x \right) = \dfrac{1}{x}$ khi $C = 0$.

Đáp án C: $\ln 2x = \ln 2 + \ln x$ là một nguyên hàm của hàm số $f\left( x \right) = \dfrac{1}{x}$ khi $C = \ln 2$.