Vấn đề 4. Tính a, b, c trong tích phân.
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Biết $\displaystyle\int\limits_1^2 \ln (9-x^2)\mathrm{\,d}x=a\ln 5+b\ln 2+c$ với $a, b, c \in \mathbb{Z}$. Tính $P=|a|+|b|+|c|$.
Các phương án chọn từ trên xuống là A B C D
$P=13$
$P=18$
$P=26$
$P=34$
Lời Giải:
Đặt $\begin{cases}&u=\ln (9-x^2)\\&\mathrm{d}v=\mathrm{d}x\end{cases} \Rightarrow \begin{cases}&\mathrm{d}u=\dfrac{-2x}{9-x^2}\mathrm{\,d}x\\&v=x+3\end{cases}.$
Khi đó $I=(x+3)\ln (9-x^2)\bigg|_1^2 +2\displaystyle\int\limits_1^2 \dfrac{x(x+3)}{9-x^2}\mathrm{\,d}x=5\ln 5-4\ln 8+2\displaystyle\int\limits_1^2 \left(-1+\dfrac{3}{3-x}\right)\mathrm{d}x$
$=5\ln 5-12\ln 2-2\left(x+3\ln |3-x|\right)\bigg|_1^2 =5\ln 5-6\ln 2-2 \to \begin{cases}a=5\\b=-6\\c=-2\end{cases} \to P=13$.
Nhận xét. Ở đây chọn $v=x+3$ thay bởi $x$ để rút gọn cho $9-x^2$, giảm thiểu biến đổi.
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Biết $\displaystyle\int\limits_0^1 \dfrac{\pi x^3+2^x+\mathrm{e}x^32^x}{\pi +\mathrm{e} \cdot 2^x}\mathrm{\,d}x=\dfrac{1}{m}+\dfrac{1}{e\ln n} \cdot \ln \left(p+\dfrac{\mathrm{e}}{\mathrm{e}+\pi}\right)$ với $m, n, p$ là các số nguyên dương. Tính tổng $P=m+n+p$.
Các phương án chọn từ trên xuống là A B C D
$P=5$
$P=6$
$P=7$
$P=8$
Lời Giải:
Ta có $I=\displaystyle\int\limits_0^1 \dfrac{\pi x^3+2^x+\mathrm{e}x^32^x}{\pi +\mathrm{e} \cdot 2^x}\mathrm{\,d}x=\displaystyle\int\limits_0^1 \left(x^3+\dfrac{2^x}{\pi +\mathrm{e} \cdot 2^x}\right)\mathrm{d}x=\dfrac{1}{4}x^4\bigg|_0^1 +A=\dfrac{1}{4}+A$.
Tính $A=\displaystyle\int\limits_0^1 \dfrac{2^x}{\pi +\mathrm{e} \cdot 2^x}\mathrm{\,d}x$. Đặt $t=\pi +\mathrm{e} \cdot 2^x \to \mathrm{\,d}t=\mathrm{e} \cdot \ln 2 \cdot 2^x\mathrm{\,d}x \to 2^x\mathrm{\,d}x=\dfrac{1}{e\ln 2}\mathrm{\,d}t$.
Đổi cận: $\begin{cases}&x=0 \to t=\pi +\mathrm{e}\\&x=1 \to t=\pi +2\mathrm{e}\end{cases}.$
Khi đó $A=\dfrac{1}{\mathrm{e} \cdot \ln 2} \cdot \displaystyle\int\limits_{\pi +\mathrm{e}}^{\pi +2\mathrm{e}} \dfrac{\mathrm{\,d}t}{t}=\dfrac{1}{\mathrm{e} \cdot \ln 2}\ln |t|\bigg|_{\pi +\mathrm{e}}^{\pi +2\mathrm{e}} =\dfrac{1}{e\ln 2}\ln \dfrac{\pi +2\mathrm{e}}{\pi +\mathrm{e}}=\dfrac{1}{e\ln 2}\ln \left(1+\dfrac{\mathrm{e}}{\mathrm{e}+\pi}\right)$.
Vậy $I=\dfrac{1}{4}+\dfrac{1}{e\ln 2}\ln \left(1+\dfrac{\mathrm{e}}{\mathrm{e}+\pi}\right) \to \begin{cases}&m=4\\&n=2\\&p=1\end{cases} \Rightarrow P=m+n+p=7$.
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Biết $\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \dfrac{x^2+\left(2x+\cos x\right)\cos x+1-\sin x}{x+\cos x}\mathrm{\,d}x=a\pi^2+b-\ln \dfrac{c}{\pi}$ với $a, b, c$ là các số hữu tỉ. Tính $P=ac^3+b$.
Các phương án chọn từ trên xuống là A B C D
$P=\dfrac{5}{4}$
$P=\dfrac{3}{2}$
$P=2$
$P=3$
Lời Giải:
Ta có $I=\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \dfrac{\left(x^2+2x\cos x+\cos^2x\right)+\left(1-\sin x\right)}{x+\cos x}\mathrm{\,d}x$
$=\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \dfrac{\left(x+\cos x\right)^2}{x+\cos x}\mathrm{\,d}x+\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \dfrac{1-\sin x}{x+\cos x}\mathrm{\,d}x=\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \left(x+\cos x\right)\mathrm{d}x+\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \dfrac{\mathrm{d}\left(x+\cos x\right)}{x+\cos x}$
$=\left(\dfrac{1}{2}x^2+\sin x+\ln \left|x+\cos x\right|\right)\bigg|_0^{\tfrac{\pi}{2}} =\dfrac{1}{8}\pi^2+1+\ln \dfrac{\pi}{2}=\dfrac{1}{8}\pi^2+1-\ln \dfrac{2}{\pi}$
$ \to \begin{cases}&a=\dfrac{1}{8}\\&b=1\\&c=2\end{cases} \to P=ac^3+b=2$.
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Biết $\displaystyle\int\limits_{\ln \sqrt{3}}^{\ln \sqrt{8}} \dfrac{1}{\sqrt{\mathrm{e}^{2x}+1}-\mathrm{e}^x}\mathrm{\,d}x=1+\dfrac{1}{2}\ln \dfrac{b}{a}+a\sqrt{a}-\sqrt{b}$ với $a, b \in \mathbb{Z}^+$. Tính $P=a+b$.
Các phương án chọn từ trên xuống là A B C D
$P=-1$
$P=1$
$P=3$
$P=5$
Lời Giải:
Ta có $I=\displaystyle\int\limits_{\ln \sqrt{3}}^{\ln \sqrt{8}} \dfrac{1}{\sqrt{\mathrm{e}^{2x}+1}-\mathrm{e}^x}\mathrm{\,d}x=\displaystyle\int\limits_{\ln \sqrt{3}}^{\ln \sqrt{8}} \left(\sqrt{\mathrm{e}^{2x}+1}+\mathrm{e}^x\right)\mathrm{d}x=\displaystyle\int\limits_{\ln \sqrt{3}}^{\ln \sqrt{8}} \sqrt{\mathrm{e}^{2x}+1}\mathrm{\,d}x+\displaystyle\int\limits_{\ln \sqrt{3}}^{\ln \sqrt{8}} \mathrm{e}^x\mathrm{\,d}x$.
$\displaystyle\int\limits_{\ln \sqrt{3}}^{\ln \sqrt{8}} \mathrm{e}^x\mathrm{\,d}x=\mathrm{e}^x\bigg|_{\ln \sqrt{3}}^{\ln \sqrt{8}} =2\sqrt{2}-\sqrt{3}$.
$\displaystyle\int\limits_{\ln \sqrt{3}}^{\ln \sqrt{8}} \sqrt{\mathrm{e}^{2x}+1}\mathrm{\,d}x$. Đặt $t=\sqrt{\mathrm{e}^{2x}+1} \Leftrightarrow t^2=\mathrm{e}^{2x}+1$, suy ra $2t\mathrm{\,d}t=2\mathrm{e}^{2x}\mathrm{\,d}x \Leftrightarrow \mathrm{\,d}x=\dfrac{t\mathrm{\,d}t}{\mathrm{e}^{2x}}=\dfrac{t\mathrm{\,d}t}{t^2-1}$.
Đổi cận: $\begin{cases}&x=\ln \sqrt{3} \to t=2\\&x=\ln \sqrt{8} \to t=3\end{cases}.$
Khi đó $\displaystyle\int\limits_{\ln \sqrt{3}}^{\ln \sqrt{8}} \sqrt{\mathrm{e}^{2x}+1}\mathrm{\,d}x=\displaystyle\int\limits_2^3 \dfrac{t^2\mathrm{\,d}t}{t^2-1}\mathrm{\,d}t=\displaystyle\int\limits_2^3 \left(1+\dfrac{1}{t^2-1}\right)\mathrm{d}t=\left(t+\dfrac{1}{2}\ln \left|\dfrac{t-1}{t+1}\right|\right)\bigg|_2^3 =1+\dfrac{1}{2}\ln \dfrac{3}{2}$.
Vậy $I=1+\dfrac{1}{2}\ln \dfrac{3}{2}+2\sqrt{2}-\sqrt{3} \to \begin{cases}&a=2\\&b=3\end{cases} \to P=a+b=5$.
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Biết $\displaystyle\int\limits_1^2 \dfrac{\mathrm{\,d}x}{(x+1)\sqrt{x}+x\sqrt{x+1}}=\sqrt{a}-\sqrt{b}-c$ với $a, b, c \in \mathbb{Z}^+$. Tính $P=a+b+c$.
Các phương án chọn từ trên xuống là A B C D
$P=12$
$P=18$
$P=24$
$P=46$
Lời Giải:
Ta có $I=\displaystyle\int\limits_1^2 \dfrac{\mathrm{\,d}x}{\sqrt{x(x+1)}\left(\sqrt{x+1}+\sqrt{x}\right)}=\displaystyle\int\limits_1^2 \dfrac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x(x+1)}\left(\sqrt{x+1}+\sqrt{x}\right)^2}\mathrm{\,d}x$.
Đặt $u=\sqrt{x+1}+\sqrt{x}$, suy ra $\mathrm{\,d}u=\left(\dfrac{1}{2\sqrt{x+1}}+\dfrac{1}{2\sqrt{x}}\right)\mathrm{d}x \to 2\mathrm{\,d}u=\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x(x+1)}}\mathrm{\,d}x$.
Đổi cận $\begin{cases}&x=2 \to u=\sqrt{3}+\sqrt{2}\\&x=1 \to u=\sqrt{2}+1\end{cases}$.
Khi đó $I=2\displaystyle\int\limits_{\sqrt{2}+1}^{\sqrt{3}+\sqrt{2}} \dfrac{\mathrm{\,d}u}{u^2}=-\dfrac{2}{u}\bigg|_{\sqrt{2}+1}^{\sqrt{3}+\sqrt{2}}=-2\left(\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{1}{\sqrt{2}+1}\right)$
$=-2\left(\dfrac{\sqrt{3}-\sqrt{2}}{3-2}-\dfrac{\sqrt{2}-1}{2-1}\right)=\sqrt{32}-\sqrt{12}-2 \to \begin{cases}&a=32\\&b=12\\&c=2\end{cases} \to P=46$.
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Biết $\displaystyle\int\limits_0^{\tfrac{\pi}{4}} \dfrac{\sin 4x}{\sqrt{\cos^2x+1}+\sqrt{\sin^2x+1}}\mathrm{\,d}x=\dfrac{a\sqrt{2}+b\sqrt{6}+c}{6}$ với $a, b, c \in \mathbb{Z}$. Tính $P=|a|+|b|+|c|$.
Các phương án chọn từ trên xuống là A B C D
$P=10$
$P=12$
$P=14$
$P=36$
Lời Giải:
Ta có $I=\displaystyle\int\limits_0^{\tfrac{\pi}{4}} \dfrac{\sin 4x}{\sqrt{\cos^2x+1}+\sqrt{\sin^2x+1}}\mathrm{\,d}x=\sqrt{2}\displaystyle\int\limits_0^{\tfrac{\pi}{4}} \dfrac{2\sin 2x\cos 2x}{\sqrt{3+\cos 2x}+\sqrt{3-\cos 2x}}\mathrm{\,d}x$.
Đặt $t=\cos 2x \to \mathrm{\,d}t=-2\sin 2x\mathrm{\,d}x$.
Đổi cận: $\begin{cases}&x=0 \to t=1\\&x=\dfrac{\pi}{4} \to t=0\end{cases}.$
Khi đó $I=-\sqrt{2}\displaystyle\int\limits_1^0 \dfrac{t}{\sqrt{3+t}+\sqrt{3-t}}\mathrm{\,d}t=\sqrt{2}\displaystyle\int\limits_0^1 \dfrac{t}{\sqrt{3+t}+\sqrt{3-t}}\mathrm{\,d}t=\dfrac{1}{\sqrt{2}}\displaystyle\int\limits_0^1 \left(\sqrt{3+t}-\sqrt{3-t}\right)\mathrm{d}t$
$=\dfrac{1}{\sqrt{2}}\left[\dfrac{2}{3}\sqrt{(3+t)^3}+\dfrac{2}{3}\sqrt{(3-t)^3}\right]\bigg|_0^1 =\dfrac{16\sqrt{2}-12\sqrt{6}+8}{6} \to \begin{cases}&a=16\\&b=-12\\&c=8\end{cases} \to P=36$.
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Biết $\displaystyle\int\limits_1^4 \sqrt{\dfrac{1}{4x}+\dfrac{\sqrt{x}+\mathrm{e}^x}{\sqrt{x}\mathrm{e}^{2x}}}\mathrm{\,d}x=a+\mathrm{e}^b-\mathrm{e}^c$ với $a, b, c \in \mathbb{Z}$. Tính $P=a+b+c$.
Các phương án chọn từ trên xuống là A B C D
$P=-5$
$P=-4$
$P=-3$
$P=3$
Lời Giải:
Ta có $\displaystyle\int\limits_1^4 \sqrt{\dfrac{1}{4x}+\dfrac{\sqrt{x}+\mathrm{e}^x}{\sqrt{x}\mathrm{e}^{2x}}}\mathrm{\,d}x=\displaystyle\int\limits_1^4 \sqrt{\dfrac{\mathrm{e}^{2x}+4x+4\mathrm{e}^x\sqrt{x}}{4x\mathrm{e}^{2x}}}\mathrm{\,d}x=\displaystyle\int\limits_1^4 \sqrt{\dfrac{\left(\mathrm{e}^x+2\sqrt{x}\right)^2}{\left(2\mathrm{e}^x\sqrt{x}\right)^2}}\mathrm{\,d}x$
$=\displaystyle\int\limits_1^4 \dfrac{\mathrm{e}^x+2\sqrt{x}}{2\mathrm{e}^x\sqrt{x}}\mathrm{\,d}x=\displaystyle\int\limits_1^4 \left(\dfrac{1}{2\sqrt{x}}+\dfrac{1}{\mathrm{e}^x}\right)\mathrm{d}x=\left(\sqrt{x}-\dfrac{1}{\mathrm{e}^x}\right)\bigg|_1^4 =1-\dfrac{1}{\mathrm{e}^4}+\dfrac{1}{\mathrm{e}}=1+\mathrm{e}^{-1}-\mathrm{e}^{-4}$
$ \to \begin{cases}&a=1\\&b=-1\\&c=-4\end{cases} \to P=a+b+c=-4$.
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Biết $\displaystyle\int\limits_0^2 \sqrt{\dfrac{2+\sqrt{x}}{2-\sqrt{x}}}\mathrm{\,d}x=a\pi +b\sqrt{2}+c$ với $a, b, c \in \mathbb{Z}$. Tính $P=a+b+c$.
Các phương án chọn từ trên xuống là A B C D
$P=-1$
$P=2$
$P=3$
$P=4$
Lời Giải:
Đặt $\sqrt{x}=2\cos u$ với $u \in \left[0; \dfrac{\pi}{2}\right]$. Suy ra $x=4\cos^2u \to \mathrm{\,d}x=-4\sin 2u\mathrm{\,d}u$.\\
Đổi cận $\begin{cases}&x=0 \to u=\dfrac{\pi}{2}\\&x=2 \to u=\dfrac{\pi}{4}\end{cases}$.
Khi đó $I=4\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \sqrt{\dfrac{2+2\cos u}{2-2\cos u}}\sin 2u\mathrm{\,d}u=8\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \dfrac{\cos \dfrac{u}{2}}{\sin \dfrac{u}{2}} \cdot \sin u \cdot \cos u\mathrm{\,d}u$
$=16\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \cos^2\dfrac{u}{2} \cdot \cos u\mathrm{\,d}u=8\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \left(1+\cos u\right) \cdot \cos u\mathrm{\,d}u=8\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \cos u\mathrm{\,d}u+4\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \left(1+\cos 2u\right)\mathrm{d}u$\\
$=8\sin u\bigg|_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} +\left(4x+2 \cdot \sin 2u\right)\bigg|_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} =\pi -4\sqrt{2}+6 \to \begin{cases}&a=1\\&b=-4\\&c=6\end{cases} \to P=3$.
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Biết $I=\displaystyle\int\limits_1^\mathrm{e} \dfrac{\ln^2x+\ln x}{\left(\ln x+x+1\right)^3}\mathrm{\,d}x=\dfrac{1}{a}-\dfrac{b}{(\mathrm{e}+2)^2}$ với $a, b \in \mathbb{Z}^+$. Tính $P=b-a$.
Các phương án chọn từ trên xuống là A B C D
$P=-8$
$P=-6$
$P=6$
$P=10$
Lời Giải:
Ta có $\displaystyle\int\limits_1^\mathrm{e} \dfrac{\ln^2x+\ln x}{\left(\ln x+x+1\right)^3}\mathrm{\,d}x=\displaystyle\int\limits_1^\mathrm{e} \dfrac{\ln x+1}{\ln x+x+1} \cdot \dfrac{\ln x}{\left(\ln x+x+1\right)^2}\mathrm{\,d}x$.
Đặt $t=\dfrac{\ln x+1}{\ln x+x+1} \to \mathrm{\,d}t=\left(\dfrac{\ln x+1}{\ln x+x+1}\right)’\mathrm{\,d}x=-\dfrac{\ln x}{\left(\ln x+x+1\right)^2}\mathrm{\,d}x$.
Đổi cận: $\begin{cases}&x=1 \to t=\dfrac{1}{2}\\&x=e \to t=\dfrac{2}{\mathrm{e}+2}\end{cases}$.
Khi đó $I=-\displaystyle\int\limits_{\tfrac{1}{2}}^{\tfrac{2}{\mathrm{e}+2}} t\mathrm{\,d}t=-\dfrac{1}{2}t^2\bigg|_{\tfrac{1}{2}}^{\tfrac{2}{\mathrm{e}+2}} =\dfrac{1}{8}-\dfrac{2}{(\mathrm{e}+2)^2}$.
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Biết $\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} \dfrac{x\cos x}{\sqrt{1+x^2}+x}\mathrm{\,d}x=a+\dfrac{\pi^2}{b}+\dfrac{\sqrt{3}\pi}{c}$ với $a, b, c$ là các số nguyên. Tính $P=a-b+c$.
Các phương án chọn từ trên xuống là A B C D
$P=-37$
$P=-35$
$P=35$
$P=41$
Lời Giải:
Ta có $I=\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} \dfrac{x\cos x}{\sqrt{1+x^2}+x}\mathrm{\,d}x=\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} x\cos x\left(\sqrt{1+x^2}-x\right)\mathrm{d}x=\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} x\left(\sqrt{1+x^2}-x\right)\cos x\mathrm{\,d}x$.
Lại có $I=\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} \dfrac{x\cos x}{\sqrt{1+x^2}+x}\mathrm{\,d}x\overset{x=-t}=\displaystyle\int\limits_{\tfrac{\pi}{6}}^{-\tfrac{\pi}{6}} \dfrac{(-t)\cos (-t)}{\sqrt{1+(-t)^2}-t}\mathrm{d}(-t)=\displaystyle\int\limits_{\tfrac{\pi}{6}}^{-\tfrac{\pi}{6}} \dfrac{t\cos t}{\sqrt{1+t^2}-t}\mathrm{\,d}t$
$=-\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} t\left(\sqrt{1+t^2}+t\right)\cos t\mathrm{\,d}t=-\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} x\left(\sqrt{1+x^2}+x\right)\cos x\mathrm{\,d}x$.
Suy ra $2I=\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} x\left(\sqrt{1+x^2}-x\right)\cos x\mathrm{\,d}x-\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} x\left(\sqrt{1+x^2}+x\right)\cos x\mathrm{\,d}x=-2\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} x^2\cos x\mathrm{\,d}x$
$ \to I=-\displaystyle\int\limits_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} x^2\cos x\mathrm{\,d}x$.
Tích phân từng phần hai lần ta được $I=2+\dfrac{\pi^2}{-36}+\dfrac{\sqrt{3}\pi}{-3}$
$\to \begin{cases}&a=2\\&b=-36\\&c=-3\end{cases} \to P=a-b+c=35$.
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