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Câu 2)
Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2\frac{\ln x}{x}dx\\ v=\frac{x^3}{3}\end{matrix}\right.\Rightarrow I=\frac{x^3}{3}\ln ^2x-\frac{2}{3}\int x^2\ln xdx\)
Đặt \(\left\{\begin{matrix} k=\ln x\\ dt=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dk=\frac{dx}{x}\\ t=\frac{x^3}{3}\end{matrix}\right.\Rightarrow \int x^2\ln xdx=\frac{x^3\ln x}{3}-\int \frac{x^2}{3}dx=\frac{x^3\ln x}{3}-\frac{x^3}{9}+c\)
Do đó \(I=\frac{x^3\ln^2x}{3}-\frac{2}{9}x^3\ln x+\frac{2}{27}x^3+c\)
Câu 3:
\(I=\int\frac{2}{\cos 2x-7}dx=-\int\frac{2}{2\sin^2x+6}dx=-\int\frac{dx}{\sin^2x+3}\)
Đặt \(t=\tan\frac{x}{2}\Rightarrow \left\{\begin{matrix} \sin x=\frac{2t}{t^2+1}\\ dx=\frac{2dt}{t^2+1}\end{matrix}\right.\)
\(\Rightarrow I=-\int \frac{2dt}{(t^2+1)\left ( \frac{4t^2}{(t^2+1)^2}+3 \right )}=-\int\frac{2(t^2+1)dt}{3t^4+10t^2+3}=-\int \frac{2d\left ( t-\frac{1}{t} \right )}{3\left ( t-\frac{1}{t} \right )^2+16}=\int\frac{2dk}{3k^2+16}\)
Đặt \(k=\frac{4}{\sqrt{3}}\tan v\). Đến đây dễ dàng suy ra \(I=\frac{-1}{2\sqrt{3}}v+c\)
\(I=\int\limits^0_{\frac{-1}{2}}\frac{dx}{\left(x+1\right)\sqrt{3+2x-x^2}}=\int\limits^0_{\frac{-1}{2}}\frac{dx}{\left(x+1\right)\left(\sqrt{\left(x+1\right)\left(3-x\right)}\right)}\)
\(=\int\limits^0_{\frac{-1}{2}}\frac{dx}{\left(x+1\right)^2\sqrt{\frac{3-x}{x+1}}}\)
Đặt \(t=\sqrt{\frac{3-x}{x+1}}\Rightarrow\frac{dx}{\left(x+1\right)^2}=-\frac{1}{2}\)
Đổi cận : \(x=-\frac{1}{2}\Rightarrow t=\sqrt{7};x=0\Rightarrow t=\sqrt{3}\)
\(I=-\frac{1}{2}\int\limits^{\sqrt{3}}_{\sqrt{7}}dt=\frac{1}{2}\left(\sqrt{7}-\sqrt{3}\right)\)
1.
Trước hết bạn nhớ công thức:
$1^2+2^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$ (cách cm ở đây: https://hoc24.vn/cau-hoi/tinh-tongs-122232n2.83618073020)
Áp vào bài:
\(\lim\frac{1}{n^3}[1^2+2^2+....+(n-1)^2]=\lim \frac{1}{n^3}.\frac{(n-1)n(2n-1)}{6}=\lim \frac{n(n-1)(2n-1)}{6n^3}\)
\(=\lim \frac{(n-1)(2n-1)}{6n^2}=\lim (\frac{n-1}{n}.\frac{2n-1}{6n})=\lim (1-\frac{1}{n})(\frac{1}{3}-\frac{1}{6n})\)
\(=1.\frac{1}{3}=\frac{1}{3}\)
2.
\(\lim \frac{1}{n}\left[(x+\frac{a}{n})+(x+\frac{2a}{n})+...+(x.\frac{(n-1)a}{n}\right]\)
\(=\lim \frac{1}{n}\left[\underbrace{(x+x+...+x)}_{n-1}+\frac{a(1+2+...+n-1)}{n} \right]\)
\(=\lim \frac{1}{n}[(n-1)x+a(n-1)]=\lim \frac{n-1}{n}(x+a)=\lim (1-\frac{1}{n})(x+a)\)
\(=x+a\)
Câu 1)
Ta có \(I=\int ^{1}_{0}\frac{dx}{\sqrt{3+2x-x^2}}=\int ^{1}_{0}\frac{dx}{4-(x-1)^2}\).
Đặt \(x-1=2\cos t\Rightarrow \sqrt{4-(x-1)^2}=\sqrt{4-4\cos^2t}=2|\sin t|\)
Khi đó:
\(I=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{d(2\cos t+1)}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{2\sin tdt}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}dt=\left.\begin{matrix} \frac{2\pi}{3}\\ \frac{\pi}{2}\end{matrix}\right|t=\frac{\pi}{6}\)
Câu 3)
\(K=\int ^{3}_{2}\ln (x^3-3x+2)dx=\int ^{3}_{2}\ln [(x+2)(x-1)^2]dx\)
\(=\int ^{3}_{2}\ln (x+2)d(x+2)+2\int ^{3}_{2}\ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\): Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln t dt=t\ln t-t\)
\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)
\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)
Bài 2)
\(J=\int ^{1}_{0}x\ln (2x+1)dx\). Đặt \(\left\{\begin{matrix} u=\ln (2x+1)\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2dx}{2x+1}\\ v=\frac{x^2}{2}\end{matrix}\right.\)
Khi đó:
\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)
\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)
\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)
1) Đặt \(2+lnx=t\Leftrightarrow x=e^{t-2}\Rightarrow dx=e^{t-2}dt\)
\(I_1=\int\left(\frac{t-2}{t}\right)^2\cdot e^{t-2}\cdot dt=\int\left(1-\frac{4}{t}+\frac{4}{t^2}\right)e^{t-2}dt\\ =\int e^{t-2}dt-4\int\frac{e^{t-2}}{t}dt+4\int\frac{e^{t-2}}{t^2}dt\)
Có:
\(4\int\frac{e^{t-2}}{t^2}dt=-4\int e^{t-2}\cdot d\left(\frac{1}{t}\right)=-\frac{4\cdot e^{t-2}}{t}+4\int\frac{e^{t-2}}{t}dt\\ \Leftrightarrow4\int\frac{e^{t-2}}{t^2}dt-4\int\frac{e^{t-2}}{t^{ }}dt=-\frac{4\cdot e^{t-2}}{t}\)
Vậy \(I_1=\int e^{t-2}dt-\frac{4\cdot e^{t-2}}{t}=e^{t-2}-\frac{4e^{t-2}}{t}+C\)
3) Đặt \(t=\sqrt{1+\sqrt[3]{x^2}}\Rightarrow t^2-1=\sqrt[3]{x^2}\Leftrightarrow x^2=\left(t^2-1\right)^3\)
\(d\left(x^2\right)=d\left[\left(t^2-1\right)^3\right]\Leftrightarrow2x\cdot dx=6t\left(t^2-1\right)^2\cdot dt\)
\(I_3=\int\frac{3t\left(t^2-1\right)^2}{t}dt=3\int\left(t^4-2t^2+1\right)dt=...\)
\(I=\int\dfrac{x}{1-cos2x}dx=\int\dfrac{x}{2sin^2x}dx\)
Đặt \(\left\{{}\begin{matrix}u=\dfrac{x}{2}\\dv=\dfrac{1}{sin^2x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{2}\\v=-cotx\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int cotxdx=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int\dfrac{cosx.dx}{sinx}\)
\(=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int\dfrac{d\left(sinx\right)}{sinx}=\dfrac{-x.cotx}{2}+\dfrac{1}{2}ln\left|sinx\right|+C\)
2/ Câu 2 bữa trước làm rồi, bạn coi lại nhé
3/ \(I=\int\left(2x+1\right)ln^2xdx\)
Đặt \(\left\{{}\begin{matrix}u=ln^2x\\dv=\left(2x+1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{2lnx}{x}dx\\v=x^2+x\end{matrix}\right.\)
\(\Rightarrow I=\left(x^2+x\right)ln^2x-\int\left(2x+2\right)lnxdx=\left(x^2+x\right)ln^2x-I_1\)
\(I_1=\int\left(2x+2\right)lnx.dx\) \(\Rightarrow\left\{{}\begin{matrix}u=lnx\\dv=\left(2x+2\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x^2+2x\end{matrix}\right.\)
\(\Rightarrow I_1=\left(x^2+2x\right)lnx-\int\left(x+2\right)dx=\left(x^2+2x\right)ln-\dfrac{x^2}{2}+2x+C\)
\(\Rightarrow I=\left(x^2+x\right)ln^2x-\left(x^2+2x\right)lnx+\dfrac{x^2}{2}-2x+C\)
4/ \(I=\int\left(2x-1\right)cosx.dx\) \(\Rightarrow\left\{{}\begin{matrix}u=2x-1\\dv=cosx.dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=sinx\end{matrix}\right.\)
\(\Rightarrow I=\left(2x-1\right)sinx-2\int sinx.dx=\left(2x-1\right)sinx+2cosx+C\)
5/ \(I=\int\left(x^2+x+1\right)e^xdx\) \(\Rightarrow\left\{{}\begin{matrix}u=x^2+x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\left(2x+1\right)dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=\left(x^2+x+1\right)e^x-\int\left(2x+1\right)e^xdx\)
\(I_1=\int\left(2x+1\right)e^xdx\) \(\Rightarrow\left\{{}\begin{matrix}u=2x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I_1=\left(2x+1\right)e^x-2\int e^xdx=\left(2x+1\right)e^x-2e^x+C=\left(2x-1\right)e^x+C\)
\(\Rightarrow I=\left(x^2+x+1\right)e^x-\left(2x-1\right)e^x+C=\left(x^2-x+2\right)e^x+C\)
6/ \(I=\int\left(2x+1\right).ln\left(x+2\right)dx\)
\(\Rightarrow\left\{{}\begin{matrix}u=ln\left(x+2\right)\\dv=\left(2x+1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x+2}\\v=x^2+x\end{matrix}\right.\)
\(\Rightarrow I=\left(x^2+x\right)ln\left(x+2\right)-\int\dfrac{x^2+x}{x+2}dx\)
\(=\left(x^2+x\right)ln\left(x+2\right)-\int\left(x-1+\dfrac{2}{x+2}\right)dx\)
\(I=\left(x^2+x\right)ln\left(x+2\right)-\dfrac{x^2}{2}+x-2ln\left|x+2\right|+C\)
Nhớ quy tắc ưu tiên khi tính nguyên hàm từng phần:
- Đặt u sẽ ưu tiên các hàm ln, log đầu tiên (luôn luôn đặt các hàm này là u nếu có mặt), sau đó đến các hàm đa thức P(x), sau đó là lượng giác hoặc e^
- Đặt dv thì theo thứ tự ngược lại, ưu tiên đặt lượng giác (sin, cos) và e^