Hàm là \(f\left(x\right)=\left(x^2-3x+3\right).e^2\) hay \(\left(x^2-3x+3\right)e^x\) bạn?
Nếu hàm là \(f\left(x\right)=\left(x^2-3x+3\right)e^2\) thì đơn giản là bạn khảo sát như khảo sát hàm \(g\left(x\right)=x^2-3x+3\)
\(f\left(x\right)=\dfrac{x^2+1}{x^3}-4x\) hay \(f\left(x\right)=\dfrac{x^2+1}{x^3-4x}\) bạn?
Trước hết ta xét: \(g\left(x\right)=\dfrac{1}{x+a}=\left(x+a\right)^{-1}\) với a là hằng số bất kì
\(g'\left(x\right)=-1.\left(x+a\right)^{-2}=\left(-1\right)^1.1!.\left(x+a\right)^{-\left(1+1\right)}\)
\(g''\left(x\right)=-1.\left(-2\right).\left(x+a\right)^{-3}=\left(-1\right)^2.2!.\left(x+a\right)^{-\left(2+1\right)}\)
Từ đó ta dễ dàng tổng quát được:
\(g^{\left(n\right)}\left(x\right)=\left(-1\right)^n.n!.\left(x+a\right)^{-\left(n+1\right)}=\dfrac{\left(-1\right)^n.n!}{\left(x+a\right)^{n+1}}\)
Xét: \(f\left(x\right)=\dfrac{x^2+1}{x\left(x-2\right)\left(x+2\right)}=-\dfrac{1}{4}.\left(\dfrac{1}{x}\right)+\dfrac{5}{8}\left(\dfrac{1}{x+2}\right)+\dfrac{5}{8}\left(\dfrac{1}{x-2}\right)\)
Áp dụng công thức trên ta được:
\(f^{\left(30\right)}\left(1\right)=\dfrac{1}{4}.\dfrac{\left(-1\right)^{30}.30!}{1^{31}}+\dfrac{5}{8}.\dfrac{\left(-1\right)^{30}.30!}{\left(1+2\right)^{31}}+\dfrac{5}{8}.\dfrac{\left(-1\right)^{30}.30!}{\left(1-2\right)^{31}}\)
Bạn tự rút gọn kết quả nhé
\(I_1=\int\left(\frac{1-cos2x}{2}\right)^2dx=\frac{1}{4}\int\left(1-2cos2x+cos^22x\right)dx\)
\(=\frac{1}{4}\int\left(1-2cos2x+\frac{1}{2}+\frac{1}{2}cos4x\right)dx\)
\(=\frac{1}{4}\int\left(\frac{3}{2}-2cos2x+\frac{1}{2}cos4x\right)dx\)
\(=\frac{1}{4}\left(\frac{3}{2}x-sin2x+\frac{1}{8}sin4x\right)+C\)
\(I_2=\int\left(\frac{1+cos6x}{2}\right)dx=\frac{1}{2}x+\frac{1}{12}sin6x+C\)
\(I_3=\int sin^2x.cos^2x.sinxdx=-\int\left(1-cos^2x\right)cos^2x.d\left(cosx\right)\)
\(=\int\left(cos^4x-cos^2x\right)d\left(cosx\right)=\frac{1}{5}cos^5x-\frac{1}{3}cos^3x+C\)
\(I_7=\int\limits^{\frac{\pi}{4}}_0\frac{1}{cos^4x}dx=\int\limits^{\frac{\pi}{4}}_0\frac{1}{cos^2x}.\frac{dx}{cos^2x}\)
Đặt \(tanx=t\Rightarrow\left\{{}\begin{matrix}\frac{1}{cos^2x}dx=dt\\x=0\Rightarrow t=0\\x=\frac{\pi}{4}\Rightarrow t=1\\\frac{1}{cos^2x}=1+tan^2x=1+t^2\end{matrix}\right.\)
\(\Rightarrow I_7=\int\limits^1_0\left(1+t^2\right)dt=\left(t+\frac{1}{3}t^3\right)|^1_0=\frac{4}{3}\)
\(I_8=\int\frac{cos^3x}{sin^3x}dx=\int\frac{cos^2x}{sin^3x}.cosxdx=\int\frac{1-sin^2x}{sin^3x}.cosxdx\)
Đặt \(sinx=t\Rightarrow cosx.dx=dt\)
\(I_8=\int\frac{1-t^2}{t^3}dt=\int\left(t^{-3}-\frac{1}{t}\right)dt=-\frac{1}{2t^2}-ln\left|t\right|+C\)
\(=-\frac{1}{2sin^2x}-ln\left|sinx\right|+C\)
\(I_9=\int\frac{\left(4x+3\right)dx}{\left(x^2-2x+2\right)^{\frac{3}{2}}}=4\int\frac{\left(x-1\right)dx}{\left(x^2-2x+2\right)^{\frac{3}{2}}}+7\int\frac{dx}{\left(x^2-2x+2\right)^{\frac{3}{2}}}=4I+7J\)
Đặt \(\sqrt{x^2-2x+2}=t\Rightarrow x^2-2x+2=t^2\)
\(\Rightarrow\left(2x-2\right)dx=2t.dt\Rightarrow\left(x-1\right)dx=t.dt\)
\(\Rightarrow I=\int\frac{tdt}{t^3}=\int\frac{1}{t^2}dt=-\frac{1}{t}+C=-\frac{1}{\sqrt{x^2-2x+2}}+C\)
Xét \(J=\int\frac{dx}{\left[\left(x-1\right)^2+1\right]^{\frac{3}{2}}}\)
Đặt \(x-1=tant\Rightarrow dx=\frac{1}{cos^2t}dt\)
\(J=\int\frac{dt}{cos^2t.\left(tan^2t+1\right)^{\frac{3}{2}}}=\int\frac{dt}{cos^2t.\left(\frac{1}{cos^2t}\right)^{\frac{3}{2}}}=\int\frac{dt}{cos^2t.\frac{1}{cos^3t}}=\int cost.dt\)
\(=sint+C\)
Mặt khác \(\left(x-1\right)^2=tan^2t=\frac{sin^2t}{1-sin^2t}\Rightarrow\frac{1}{\left(x-1\right)^2}=\frac{1}{sin^2t}-1\)
\(\Rightarrow\frac{1}{sin^2t}=\frac{1}{\left(x-1\right)^2}+1=\frac{x^2-2x+2}{\left(x-1\right)^2}\Rightarrow\frac{1}{sint}=\frac{\sqrt{x^2-2x+2}}{x-1}\)
\(\Rightarrow sint=\frac{x-1}{\sqrt{x^2-2x+2}}\)
\(\Rightarrow J=\frac{x-1}{\sqrt{x^2-2x+2}}+C\)
\(\Rightarrow I_9=-\frac{4}{\sqrt{x^2-2x+2}}+\frac{7\left(x-1\right)}{\sqrt{x^2-2x+2}}+C=\frac{7x-11}{\sqrt{x^2-2x+2}}+C\)
\(\Leftrightarrow\left[f'\left(x\right)\right]^2+f\left(x\right).f''\left(x\right)=\frac{1}{x^3}\)
\(\Leftrightarrow\left[f'\left(x\right).f\left(x\right)\right]'=\frac{1}{x^3}\)
Lấy nguyên hàm 2 vế:
\(f'\left(x\right).f\left(x\right)=\int\frac{1}{x^3}dx=-\frac{1}{2x^2}+C\)
Thay \(x=\frac{1}{4}\Rightarrow-8=-8+C\Rightarrow C=0\)
\(\Rightarrow f'\left(x\right)f\left(x\right)=-\frac{1}{2x^2}\)
Tiếp tục lấy nguyên hàm 2 vế:
\(\int f\left(x\right).f'\left(x\right)dx=-\frac{1}{2}\int\frac{1}{x^2}dx\)
\(\Leftrightarrow\frac{1}{2}f^2\left(x\right)=\frac{1}{2x}+C\)
Thay \(x=\frac{1}{4}\Rightarrow2=2+C\Rightarrow C=0\Rightarrow f^2\left(x\right)=\frac{1}{x}\)
\(\Rightarrow f\left(x\right)=\frac{1}{\sqrt{x}}\Rightarrow f'\left(x\right)=-\frac{1}{2x\sqrt{x}}\)
\(\Rightarrow m=\int\limits^{16}_1-\frac{1}{\frac{2x}{4}\sqrt{\frac{x}{4}}}dx=-6\)
Không có cách nào nhanh được cả, chỉ có kiên nhẫn nguyên hàm từng phần thôi bạn
Sẽ tách ra và nguyên hàm từng phần lần lượt từ mũ to đến mũ nhỏ:
\(I=\int\frac{-2ln^3x+3ln^2x+11lnx-2}{x^2}dx=\int\frac{-2ln^3x}{x^2}dx+\int\frac{3ln^2x+11lnx-2}{x^2}dx\)
Đặt \(\left\{{}\begin{matrix}u=-2ln^3x\\dv=\frac{1}{x^2}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{-6ln^2x}{x}dx\\v=-\frac{1}{x}\end{matrix}\right.\)
\(\Rightarrow I=\frac{2ln^3x}{x}-\int\frac{6ln^2x}{x^2}dx+\int\frac{3ln^2x+11lnx-2}{x^2}dx\)
Tiếp tục tính cho bậc 2 cụm \(-\frac{6ln^2x}{x^2}+\frac{3ln^2x}{x^2}=-\frac{3ln^2x}{x^2}\)
Đặt \(\left\{{}\begin{matrix}u=-3ln^2x\\dv=\frac{1}{x^2}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-\frac{6lnx}{x}dx\\v=-\frac{1}{x}\end{matrix}\right.\)
\(\Rightarrow\frac{3ln^2x}{x}-\int\frac{6lnx}{x^2}\)
Lại gom và tính tiếp phần \(\frac{lnx}{x^2}\) ...
Sau 4 lần tính như vậy sẽ ra kết quả thôi
Chỗ lấy nguyên hàm của \(\int\frac{-2ln^3x+3ln^2x+11lnx-2}{x^2}dx\) bạn giải rõ giúp mình được không
\(\left[f'\left(x\right)\right]^2+f\left(x\right).f''\left(x\right)=\frac{-2ln^3x+3ln^2x+11lnx-2}{x^2}\)
\(\Leftrightarrow\left[f\left(x\right).f'\left(x\right)\right]'=\frac{-2ln^3x+3ln^2x+11lnx-2}{x^2}\)
Lấy nguyên hàm 2 vế:
\(f\left(x\right).f'\left(x\right)=\int\frac{-2ln^3x+3ln^2x+11lnx-2}{x^2}dx=\frac{2ln^3x+3ln^2x-5lnx-3}{x}+C\)
Thay \(x=\frac{1}{e}\Rightarrow3e=3e+C\Rightarrow C=0\)
\(\Rightarrow f\left(x\right).f'\left(x\right)=\frac{2ln^3x+3ln^2x-5lnx-3}{x}\)
Tiếp tục lấy nguyên hàm 2 vế:
\(\Leftrightarrow\frac{1}{2}f^2\left(x\right)=\int\frac{2ln^3x+3ln^2x-5lnx-3}{x}dx=\frac{1}{2}\left(ln^4x+2ln^3x-5ln^2x-6lnx\right)+C\)
Thay \(x=\frac{1}{e}\Rightarrow\frac{9}{2}=C\Rightarrow C=\frac{9}{2}\)
\(\Rightarrow f^2\left(x\right)=ln^4x+2ln^3x-5ln^2x-6lnx+9\)
\(\Rightarrow f\left(x\right)=\sqrt{ln^4x+2ln^3x-5ln^2x-6lnx+9}\)
\(\Rightarrow m=f\left(e\right)-f\left(\frac{1}{e}\right)=1-3=-2\)
\(I_1=\int\left(\frac{6}{2\sqrt{x}}+2\sqrt{x}\right)dx=6\sqrt{x}+\frac{4}{3}\sqrt{x^3}+C\)
\(I_2=\int\frac{xdx}{\sqrt{5-4x^2}}\)
Đặt \(\sqrt{5-4x^2}=t\Rightarrow5-4x^2=t^2\)
\(\Rightarrow-8xdx=2tdt\Rightarrow xdx=-\frac{1}{4}tdt\)
\(\Rightarrow I_2=\int\frac{-\frac{1}{4}t.dt}{t}=-\frac{1}{4}\int dt=-\frac{1}{4}t+C=-\frac{1}{4}\sqrt{5-4x^2}+C\)
\(I_3=\int\frac{dx}{\left(2x-3\right)^2}=\frac{1}{2}\int\frac{d\left(2x-3\right)}{\left(2x-3\right)^2}=-\frac{1}{2\left(2x-3\right)}+C\)
\(I_4=\int\frac{\left(1+\sqrt{x}\right)^{\frac{1}{4}}}{\sqrt{x}}dx\)
Đặt \(\left(1+\sqrt{x}\right)^{\frac{1}{4}}=t\Rightarrow\sqrt{x}=t^4-1\)
\(\Rightarrow x=t^8-2t^4+1\Rightarrow dx=\left(8t^7-8t^3\right)dt\)
\(\Rightarrow I_4=\int\frac{t\left(8t^7-8t^3\right)}{t^4-1}dt=\int\frac{8t^4\left(t^4-1\right)}{t^4-1}dt=\int8t^4dt\)
\(=\frac{8}{5}t^5+C=\frac{8}{5}\left(1+\sqrt{x}\right)^{\frac{5}{4}}+C\)
\(I_5=\frac{2+3x}{\sqrt{1+4x+3x^2}}dx\)
Đặt \(\sqrt{1+4x+3x^2}=t\Rightarrow1+4x+3x^2=t^2\)
\(\Leftrightarrow\left(4+6x\right)dx=2tdt\Rightarrow\left(2+3x\right)dx=t.dt\)
\(\Rightarrow I_5=\int\frac{t.dt}{t}=\int dt=t+C=\sqrt{1+4x+3x^2}+C\)
\(I_6=\int\frac{lnx.dx}{x\left(1+ln^2x\right)}\)
Đặt \(1+ln^2x=t\Rightarrow\frac{2lnx}{x}dx=dt\Rightarrow\frac{lnxdx}{x}=\frac{dt}{2}\)
\(\Rightarrow I_6=\int\frac{dt}{2t}=\frac{1}{2}ln\left|t\right|+C=\frac{1}{2}ln\left(1+ln^2x\right)+C\)
\(I_7=\int\frac{e^{tan^{-1}x}dx}{1+x^2}\)
Đặt \(tan^{-1}x=t\Rightarrow\frac{dx}{1+x^2}=dt\)
\(\Rightarrow I_7=\int e^tdt=e^t+C=e^{tan^{-1}x}+C\)
\(I_8=\int\frac{sin\sqrt{x}}{\sqrt{x}}dx\)
Đặt \(\sqrt{x}=t\Rightarrow\frac{dx}{2\sqrt{x}}=dt\Rightarrow\frac{dx}{\sqrt{x}}=2dt\)
\(\Rightarrow I_8=\int2sint.dt=-2cost+C=-2cos\left(\sqrt{x}\right)+C\)
\(\left[\left(2x+1\right)^4-2\left(2x+1\right)^2-1\right]f'\left(x\right)=\left(64x^3+96x^2+32x\right)f\left(x\right)\)
\(\Leftrightarrow\frac{f'\left(x\right)}{f\left(x\right)}=\frac{64x^3+96x^2+32x}{\left(2x+1\right)^4-2\left(2x+1\right)^2-1}=\frac{32x\left(x+1\right)\left(2x+1\right)}{\left(2x+1\right)^4-2\left(2x+1\right)^2-1}\)
Lấy nguyên hàm 2 vế:
\(\int\frac{f'\left(x\right)}{f\left(x\right)}dx=\int\frac{32x\left(x+1\right)\left(2x+1\right)}{\left(2x+1\right)^4-2\left(2x+1\right)^2-1}dx\)
\(\Leftrightarrow ln\left|f\left(x\right)\right|=\int\frac{32x\left(x+1\right)\left(2x+1\right)}{\left(2x+1\right)^4-2\left(2x+1\right)^2-1}dx\)
Xét \(I=\int\frac{32x\left(x+1\right)\left(2x+1\right)}{\left(2x+1\right)^4-2\left(2x+1\right)^2-1}dx=\int\frac{4\left(2x+1-1\right)\left(2x+1+1\right)\left(2x+1\right)}{\left(2x+1\right)^4-2\left(2x+1\right)^2-1}d\left(2x+1\right)\)
Đặt \(t=2x+1\Rightarrow I=\int\frac{4\left(t-1\right)\left(t+1\right)t}{t^4-2t^2-1}dt=\int\frac{4t^3-4t}{t^4-2t^2-1}dt\)
\(=\int\frac{d\left(t^4-2t^2-1\right)}{t^4-2t^2-1}=ln\left|t^4-2t^2-1\right|+C\)
\(=ln\left|\left(2x+1\right)^4-2\left(2x+1\right)^2-1\right|+C\)
\(\Rightarrow ln\left|f\left(x\right)\right|=ln\left|\left(2x+1\right)^4-2\left(2x+1\right)^2-1\right|+C\)
Thay \(x=1\Rightarrow ln\left(62\right)=ln\left(62\right)+C\Rightarrow C=0\)
\(\Rightarrow ln\left|f\left(x\right)\right|=ln\left[\left(2x+1\right)^4-2\left(2x+1\right)^2-1\right]\)
\(\Rightarrow f\left(x\right)=-\left(2x+1\right)^4+2\left(2x+1\right)^2+1\)
\(f\left(2\right)=-574\)
Đề đúng chứ bạn?
Biểu thức \(-\left(2x+1\right)^4+2\left(2x+1\right)^2+1\) chắc chắn đúng chứ?
\(\Leftrightarrow\frac{f'\left(x\right)}{f\left(x\right)}=\frac{2\sqrt{x+4}+1}{2\sqrt{x+4}\left(\sqrt{x+4}+x+5\right)}\)
Lấy nguyên hàm 2 vế: \(\int\frac{f'\left(x\right)}{f\left(x\right)}dx=\int\frac{2\sqrt{x+4}+1}{2\sqrt{x+4}\left(\sqrt{x+4}+x+5\right)}dx\)
\(\Leftrightarrow ln\left(f\left(x\right)\right)=\int\frac{2\sqrt{x+4}+1}{2\sqrt{x+4}\left(\sqrt{x+4}+x+5\right)}dx\)
Xét \(I=\int\frac{2\sqrt{x+4}+1}{2\sqrt{x+4}}.\frac{1}{\left(\sqrt{x+4}+x+5\right)}dx\)
Đặt \(\sqrt{x+4}+x+5=t\)
\(\Rightarrow\left(\frac{1}{2\sqrt{x+4}}+1\right)dx=dt\Leftrightarrow\frac{2\sqrt{x+4}+1}{2\sqrt{x+4}}dx=dt\)
\(\Rightarrow I=\int\frac{dt}{t}=lnt+C=ln\left(x+5+\sqrt{x+4}\right)+C\)
\(\Rightarrow ln\left(f\left(x\right)\right)=ln\left(x+5+\sqrt{x+4}\right)+C\)
Thế \(x=-3\) vào biểu thức trên:
\(ln\left(3\right)=ln\left(2+1\right)+C\Leftrightarrow C=0\)
\(\Leftrightarrow ln\left(f\left(x\right)\right)=ln\left(x+5+\sqrt{x+4}\right)\)
\(\Rightarrow f\left(x\right)=x+5+\sqrt{x+4}\)
\(\Rightarrow f\left(5\right)=13\)