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1)
\(\int\frac{tan^3x}{cos2x}dx=\int\frac{sin^3x}{cos^3x\cdot\left(2cos^2x-1\right)}dx=\int\frac{1-cos^2x}{cos^3x\left(2cos^2x-1\right)}\cdot sinx\cdot dx\\ =\int\frac{1-cos^2x}{cos^3x\left(2cos^2x-1\right)}d\left(cosx\right)=...\)
Câu 1)
\(I=\int \ln ^3 xdx\). Đặt \(\left\{\begin{matrix} u=\ln ^3x\\ dv=dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{3\ln ^2x}{x}dx\\ v=x\end{matrix}\right.\)
\(\Rightarrow I=x\ln ^3x-3\int \ln^2xdx\)
Tiếp tục nguyên hàm từng phần cho \(\int \ln ^2xdx\) như trên, ta suy ra:
\(\int\ln ^2xdx=x\ln^2x-2\int \ln x dx\).
Tiếp tục nguyên hàm từng phần cho \(\int \ln xdx\Rightarrow \int \ln xdx=x\ln x-x+c\)
Do đó mà \(I=x\ln ^3x-3(x\ln^2x-2x\ln x+2x)+c\)
\(\Leftrightarrow I=x\ln^3x-3x\ln^2x+6x\ln x-6x+c\)
Câu 2)
\(I=\int ^{1}_{0}(x+\sin ^2x)\cos x dx=\int ^{1}_{0}x\cos xdx+\int ^{1}_{0}\sin^2x\cos xdx\)
Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\sin x\end{matrix}\right.\Rightarrow \int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+c\)
\(\Rightarrow \int ^{1}_{0} x\cos xdx=\sin 1+\cos 1-1\)
Còn \(\int ^{1}_{0}\sin^2x\cos xdx=\int ^{1}_{0}\sin ^2xd(\sin x)=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\sin ^3x}{3}=\frac{\sin^31}{3}\)
\(\Rightarrow I=-1+\sin 1+\cos 1+\frac{\sin ^3 1}{3}\approx 0,0173\)
a) Đặt \(\sqrt{2x-5}=t\) khi đó \(x=\frac{t^2+5}{2}\) , \(dx=tdt\)
Do vậy \(I_1=\int\frac{\frac{1}{4}\left(t^2+5\right)^2+3}{t^3}dt=\frac{1}{4}\int\frac{\left(t^4+10t^2+37\right)t}{t^3}dt\)
\(=\frac{1}{4}\int\left(t^2+10+\frac{37}{t^2}\right)dt=\frac{1}{4}\left(\frac{t^3}{3}+10t-\frac{37}{t}\right)+C\)
Trở về biến x, thu được :
\(I_1=\frac{1}{12}\sqrt{\left(2x-5\right)^3}+\frac{5}{2}\sqrt{2x-5}-\frac{37}{4\sqrt{2x-5}}+C\)
b) \(I_2=\frac{1}{3}\int\frac{d\left(\ln\left(3x-1\right)\right)}{\ln\left(3x-1\right)}=\frac{1}{3}\ln\left|\ln\left(3x-1\right)\right|+C\)
c) \(I_3=\int\frac{1+\frac{1}{x^2}}{\sqrt{x^2-7+\frac{1}{x^2}}}dx=\int\frac{d\left(x-\frac{1}{x}\right)}{\sqrt{\left(x-\frac{1}{2}\right)^2-5}}\)
Đặt \(x-\frac{1}{x}=t\)
\(\Rightarrow\) \(I_3=\int\frac{dt}{\sqrt{t^2-5}}=\ln\left|t+\sqrt{t^2-5}\right|+C\)
\(=\ln\left|x-\frac{1}{x}+\sqrt{x^2-7+\frac{1}{x^2}}\right|+C\)
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 \(t=\sqrt{1+\sqrt[3]{x^2}}\Rightarrow t^2-1=\sqrt[3]{x^2}\Leftrightarrow x^2=\left(t^2-1\right)^3\)
\(d\left(x^2\right)=d\left[\left(t^2-1\right)^3\right]\Leftrightarrow2x\cdot dx=6t\left(t^2-1\right)^2\cdot dt\)
\(I_3=\int\frac{3t\left(t^2-1\right)^2}{t}dt=3\int\left(t^4-2t^2+1\right)dt=...\)
1/ \(\int\limits^e_1\left(x+\frac{1}{x}+\frac{1}{x^2}\right)dx=\left(\frac{x^2}{2}+lnx-\frac{1}{x}\right)|^e_1=\frac{e^2}{2}-\frac{1}{e}+\frac{3}{2}\)
2/ \(\int\limits^2_1\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)dx=\int\limits^2_1\left(x\sqrt{x}+1\right)dx=\int\limits^2_1\left(x^{\frac{3}{2}}+1\right)dx\)
\(=\left(\frac{2}{5}.x^{\frac{5}{2}}+x\right)|^2_1=\frac{8\sqrt{2}-7}{5}\)
3/
\(\int\limits^2_1\frac{2x^3-4x+5}{x}dx=\int\limits^2_1\left(2x^2-4+\frac{5}{x}\right)dx=\left(\frac{2}{3}x^3-4x+5lnx\right)|^2_1=\frac{2}{3}+5ln2\)
4/ \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx=\int\limits^2_1\left(6x^2-2x\right)dx=\left(2x^3-x^2\right)|^2_1=11\)
a/ Tích phân này làm sao giải được nhỉ?
b/ Đặt \(\sqrt{x}=t\Rightarrow x=t^2\Rightarrow dx=2t.dt\)
\(I=\int\frac{2t^2.dt}{4-t^4}=\int\left(\frac{1}{2-t^2}-\frac{1}{2+t^2}\right)dt=\frac{1}{2\sqrt{2}}ln\left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{\sqrt{2}}arctan\frac{\sqrt{2}}{t}+C\)
\(=\frac{1}{2\sqrt{2}}ln\left|\frac{\sqrt{2}+\sqrt{x}}{\sqrt{2}-\sqrt{x}}\right|+\frac{1}{\sqrt{2}}arctan\frac{\sqrt{2}}{\sqrt{x}}+C\)
c/ \(I=\int\frac{\sqrt{1+x^2}}{x^2}.xdx\)
Đặt \(\sqrt{1+x^2}=t\Rightarrow x^2=t^2-1\Rightarrow xdx=tdt\)
\(\Rightarrow I=\int\frac{t^2dt}{t^2-1}=\int\left(1+\frac{1}{t^2-1}\right)dt=t+ln\left|\frac{t-1}{t+1}\right|+C=\sqrt{1+x^2}+ln\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C\)
d/ Con nguyên hàm này cũng không tính được, chắc bạn ghi nhầm đề
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:
\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)
\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)
\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)
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 \(I=\frac{-1}{2\sqrt{3}}v+c\)
Câu 6)
\(I=-\int \frac{\left ( 1-\frac{1}{x^2} \right )dx}{x^2+2+\frac{1}{x^2}}=-\int \frac{d\left ( x+\frac{1}{x} \right )}{\left ( x+\frac{1}{x} \right )^2}=-\frac{1}{x+\frac{1}{x}}+c=-\frac{x}{x^2+1}+c\)
Câu 8)
\(I=\int \ln \left(\frac{x+1}{x-1}\right)dx=\int \ln (x+1)dx-\int \ln (x-1)dx\)
\(\Leftrightarrow I=\int \ln (x+1)d(x+1)-\int \ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\) ta có:
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)
\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)
\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)
Câu 5)
\(I=\int \frac{1+x(2\ln x-1)}{x(x+1)^2}dx=\int \frac{dx}{x(x+1)^2}-\int \frac{dx}{(x+1)^2}+\int \frac{2\ln xdx}{(x+1)^2}\)
\(=\int \left ( \frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1)^2} \right )dx+\int \frac{2\ln xdx}{(x+1)^2}-\int \frac{dx}{(x+1)^2}\)
\(\Leftrightarrow I=\ln|x|-\ln|x+1|+\frac{2}{x+1}+\int \frac{2\ln x dx}{(x+1)^2}+c\)
Đối với \(\int \frac{2\ln x}{(x+1)^2}dx\), ta đặt \(\left\{\begin{matrix} u=\ln x\\ dv=\frac{dx}{(x+1)^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dx}{x}\\ v=\frac{-1}{x+1}\end{matrix}\right.\)
\( \Rightarrow \int \frac{2\ln xdx}{(x+1)^2}=\frac{-2\ln x}{x+1}+\int \frac{1}{x(x+1)}dx=\frac{-2\ln x}{x+1}+2\ln |x|-2\ln|x+1|+c\)
\(\Rightarrow I=3\ln |x|-3\ln|x+1|+\frac{2}{x+1}-\frac{2\ln x}{x+1}+c\)
Lời giải
Đặt \(x=2t\). Biến đổi ta có:
\(I=\int e^{2t}\frac{(\sin t+\cos t)^2}{\cos^2t}dt=\int e^{2t}(\tan t+1)^2dt\)
\(\Leftrightarrow I=\int e^{2t}\tan^2t dt+2\int e^{2t}\tan tdt+\int e^{2t}dt\)
Đặt \(\left\{\begin{matrix} u=e^{2t}\\ dv=\tan^2tdt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2e^{2t}dt\\ v=\int \tan^2tdt=\int \frac{1-\cos^2t}{\cos^2t}dt=\tan t-t\end{matrix}\right.\)
\(\Rightarrow I=e^{2t}(\tan t-t)-2\int (\tan t-t)e^{2t}dt+2\int e^{2t}\tan tdt+\int e^{2t}dt\)
\(\Leftrightarrow I=e^{2t}(\tan t-t)+2\int e^{2t}tdt+\int e^{2t}dt\)
Đặt \(\left\{\begin{matrix} k=t\\ dl=e^{2t}dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dk=dt\\ l=\int e^{2t}dt=\frac{e^{2t}}{2}\end{matrix}\right.\)
\(\Rightarrow I=e^{2t}(\tan t-t)+te^{2t}-\int e^{2t}dt+\int e^{2t}dt\)
Hay \(I=e^{2t}\tan t+c=e^x\tan \frac{x}{2}+c\)
Câu 11)
\(I=\int \frac{x^3\ln x}{\sqrt{x^2+1}}=\int \frac{x^2\ln (x^2)d(x^2)}{4\sqrt{x^2+1}}\). Đặt \(\sqrt{x^2+1}=t\rightarrow x^2=t^2-1\).
Khi đó \(I=\int \frac{(t^2-1)\ln(t^2-1)d(t^2-1)}{4t}=\frac{1}{2}\int (t^2-1)\ln (t^2-1)dt\)
Đặt \(\left\{\begin{matrix} u=\ln (t^2-1)\\ dv=(t^2-1)dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2t}{t^2-1}dt\\ v=\frac{t^3}{3}-t\end{matrix}\right.\)
\(\Rightarrow 2I=\ln (t^2-1)\left(\frac{t^3}{3}-t\right)-2\int \left(\frac{t^3}{3}-t\right)\frac{t}{t^2-1}dt\)
Đối với \(\int \frac{t^4}{t^2-1}dt\)
\(\int \frac{t^4}{t^2-1}dt=\int (t^2+1+\frac{1}{t^2-1})dt=\frac{t^3}{3}+t+\frac{\ln|t-1|-\ln|t+1|}{2}+c\)
Đối với \(\int \frac{t^2}{t^2-1}dt\)
\(\int \frac{t^2}{t^2-1}dt=\int (1+\frac{1}{t^2-1})dt=t+\frac{\ln|t-1|-\ln|t+1|}{2}+c\)
Do đó mà \(I=\frac{1}{2}\left[\left(\frac{t^3}{3}-t\right)\ln (t^2-1)-\frac{2t^3}{9}+\frac{4t}{3}+\frac{2(\ln|t-1|-\ln|t+1|)}{3}\right]+c\)
Câu 10.
Đặt \(x-1=t\Rightarrow I=\int \frac{\ln t}{t^4}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^4}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{3t^3}\end{matrix}\right.\Rightarrow I=\frac{-\ln t}{3t^3}+\int \frac{dt}{3t^4}=\frac{-\ln t}{3t^3}-\frac{1}{9t^3}+c\)
Câu 12)
Đặt \(\left\{\begin{matrix} u=x\\ dv=\frac{e^x}{(e^x+1)^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\int \frac{e^x}{(e^x+1)^2}dx=\int\frac{d(e^x+1)}{(e^x+1)^2}=-\frac{1}{e^x+1}\end{matrix}\right.\)
\(\Rightarrow I=\frac{-x}{e^x+1}+\int \frac{dx}{e^x+1}=\frac{-x}{e^x+1}+\int \frac{d(e^x)}{e^x(e^x+1)}=\frac{-x}{e^x+1}+\ln|e^x|-\ln|e^x+1|+c\)
Hay \(I=\frac{xe^x}{e^x+1}-\ln|e^x+1|+c\)
Câu 3)
\(I=\int \frac{3x\cos x+2}{1+\cot ^2x}dx=\int (3x\cos x+2)\sin^2xdx=3\int x\cos x\sin^2xdx+2\int sin^2xdx \)
Ta có:
\(2\int \sin^2xdx=\int (1-\cos 2x)dx=x-\frac{\sin 2x}{2}+c\)
Đối với \(3\int x\cos x\sin^2xdx\): Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos x\sin^2xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\int \sin^2xd(\sin x)=\frac{\sin^3x}{3}\end{matrix}\right.\)
\(\Rightarrow 3\int x\cos x\sin^2xdx=x\sin^3x-\int \sin^3xdx=x\sin^3x-\int \frac{3\sin x-\sin 3x}{4}dx\)
\(=x\sin^3x+\frac{3}{4}\cos x-\frac{\cos 3x}{12}+c\)
Câu 7 nhé.
Akai ơi, bà cho tôi giải ké 1 câu với :)))
1)
\(ln\frac{\left(1+sinx\right)^{1+cosx}}{1+cosx}\\ =\left(1+cosx\right)ln\left(1+sinx\right)-ln\left(1+cosx\right)\\ =\left[ln\left(1+sinx\right)-ln\left(1+cosx\right)\right]+cosx\cdot ln\left(1+sinx\right)\\ =ln\frac{1+sinx}{1+cosx}+cosx\cdot ln\left(1+sinx\right)\)
Ta có: \(I=\int ln\frac{1+sinx}{1+cosx}dx+\int cosx\cdot ln\left(1+sinx\right)dx\)
+) \(B=\int cosx\cdot ln\left(1+sinx\right)dx\\ =\int ln\left(1+sinx\right)d\left(1+sinx\right)=\int ln\left(t\right)dt=...\)
+) \(A=\int ln\frac{1+sinx}{1+cosx}dx\)
Đặt: \(f\left(x\right)=\frac{1+sinx}{1+cosx}\Rightarrow f\left(t\right)=\frac{1+sint}{1+cost}\)
Hoàn toàn có thể đặt: \(f\left(x\right)=e^{f\left(t\right)}\)
Đạo hàm 2 vế ta có: \(f'\left(x\right)dx=e^{f\left(t\right)}.e^t.dt\)
Ta có: \(A=\int ln\left(e^{f\left(t\right)}\right).e^t.dt=\int e^t.f\left(t\right)dt=\int e^t.\frac{1+sint}{1+cost}dt\)
(đến đây thì làm giống câu 7 thôi)
em chịu
;v Giải từng câu để dc nhìu GP :O
cc cc ko hiểu j cả ngu vkl!