Chương 3: NGUYÊN HÀM. TÍCH PHÂN VÀ ỨNG DỤNG - Hỏi đáp

a/ \(I=\int\limits^1_0\dfrac{1}{\left(x^2+3\right)\left(x^2+1\right)}dx=\dfrac{1}{2}\int\limits^1_0\left(\dfrac{1}{x^2+1}-\dfrac{1}{x^2+3}\right)dx\)

\(=\dfrac{1}{2}\left(arctanx-\dfrac{1}{\sqrt{3}}arctan\dfrac{x}{\sqrt{3}}\right)|^1_0=\dfrac{\pi}{8}-\dfrac{\pi\sqrt{3}}{36}\)

b/ \(I=\int\dfrac{x^2-1}{x^4+1}dx=\int\dfrac{1-\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}dx\)

Đặt \(x+\dfrac{1}{x}=t\Rightarrow\left(1-\dfrac{1}{x^2}\right)dx=dt\) ; \(x^2+\dfrac{1}{x^2}=t^2-2\)

\(\Rightarrow I=\int\dfrac{dt}{t^2-2}=\int\dfrac{dt}{\left(t-\sqrt{2}\right)\left(t+\sqrt{2}\right)}=\dfrac{1}{2\sqrt{2}}\int\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)dt\)

\(\Rightarrow I=\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{t-\sqrt{2}}{t+\sqrt{2}}\right|+C=\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|+C\)

c/ \(I=\int\dfrac{dx}{x\left(x^3+1\right)}=\int\dfrac{x^2dx}{x^3\left(x^3+1\right)}\)

Đặt \(x^3+1=t\Rightarrow3x^2dx=dt\)

\(\Rightarrow I=\dfrac{1}{3}\int\dfrac{dt}{\left(t-1\right)t}=\dfrac{1}{3}\int\left(\dfrac{1}{t-1}-\dfrac{1}{t}\right)dt=\dfrac{1}{3}ln\left|\dfrac{t-1}{t}\right|+C\)

\(\Rightarrow I=\dfrac{1}{3}ln\left|\dfrac{x^3}{x^3+1}\right|+C\)

d/ \(I=\int\limits^1_0\dfrac{xdx}{x^4+x^2+1}\)

Đặt \(x^2=t\Rightarrow2xdx=dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=1\Rightarrow t=1\end{matrix}\right.\)

\(I=\dfrac{1}{2}\int\limits^1_0\dfrac{dt}{t^2+t+1}=\dfrac{1}{2}\int\limits^1_0\dfrac{dt}{\left(t+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}=\dfrac{2}{3}\int\limits^1_0\dfrac{dt}{\dfrac{4}{3}\left(t+\dfrac{1}{2}\right)^2+1}\)

Đặt \(t+\dfrac{1}{2}=\dfrac{\sqrt{3}}{2}tanu\Rightarrow dt=\dfrac{\sqrt{3}}{2}.\dfrac{du}{cos^2u}\); \(\left\{{}\begin{matrix}t=0\Rightarrow u=\dfrac{\pi}{6}\\t=1\Rightarrow u=\dfrac{\pi}{3}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{2}{3}.\dfrac{\sqrt{3}}{2}\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{du}{cos^2u\left(tan^2u+1\right)}=\dfrac{\sqrt{3}}{3}\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}du=\dfrac{\pi\sqrt{3}}{18}\)

giup minh voi

\(b,I=\int tanxdx=\int\dfrac{sinx}{cosx}dx\)
Đặt u= cosx => du= -sinxdx
\(I=\int-\dfrac{1}{u}du=-\int u^{-1}du=-ln\left|u\right|+C=-lncosx+C\)

\(g,\int\dfrac{1}{sin^2x.cos^2x}=\int\dfrac{sin^2x+cos^2x}{sin^2x.cos^2x}=\int\dfrac{1}{cos^2x}dx+\int\dfrac{1}{sin^2x}dx=tanx-cotx+C\)

Nếu miền lấy tích phân không sai, ví dụ sửa thành tính tích phân trên miền \(\left[0;\dfrac{1}{2}\right]\) thì bạn cứ phá trị tuyệt đối rồi làm như bình thường thôi, ko có gì đặc biệt cả:

\(I=\int\limits^{\dfrac{1}{2}}_0\dfrac{\left|4x-1\right|}{x^2-3x+2}dx=\int\limits^{\dfrac{1}{4}}_0\dfrac{1-4x}{x^2-3x+2}dx+\int\limits^{\dfrac{1}{2}}_{\dfrac{1}{4}}\dfrac{4x-1}{x^2-3x+2}dx=I_1+I_2\)

\(I_1=\int\limits^{\dfrac{1}{4}}_0\dfrac{-2\left(2x-3\right)-5}{x^2-3x+2}dx=-2\int\limits^{\dfrac{1}{4}}_0\dfrac{2x-3}{x^2-3x+2}dx-5\int\limits^{\dfrac{1}{4}}_0\dfrac{1}{x^2-3x+2}dx\)

\(=-2\int\limits^{\dfrac{1}{4}}_0\dfrac{d\left(x^2-3x+2\right)}{x^2-3x+2}-5\int\limits^{\dfrac{1}{4}}_0\dfrac{1}{\left(x-1\right)\left(x-2\right)}dx\)

\(=-2ln\left|x^2-3x+2\right|^{\dfrac{1}{4}}_0-5ln\left|\dfrac{x-2}{x-1}\right|^{\dfrac{1}{4}}_0=...\)

Tương tự:

\(I_2=\int\limits^{\dfrac{1}{2}}_{\dfrac{1}{4}}\dfrac{4x-1}{x^2-3x+2}dx=\int\limits^{\dfrac{1}{2}}_{\dfrac{1}{4}}\dfrac{2\left(3x-2\right)+5}{x^2-3x+2}dx=...\)

\(\Rightarrow I=I_1+I_2=...\)

Đơn giản là đề sai bạn ơi, trên đoạn lấy tích phân có tới 2 điểm mà hàm số không xác định

Muốn lấy tích phân thì trước hết ở đoạn lấy tích phân, hàm số phải liên tục và xác định đã (tất nhiên là không nói đến tích phân suy rộng loại 2 của toán cao cấp)

\(\int\dfrac{1}{1+sinx}dx=\int\dfrac{1-sinx}{\left(1-sinx\right)\left(1+sinx\right)}dx=\int\dfrac{1-sinx}{1-sin^2x}dx\)

\(=\int\dfrac{1-sinx}{cos^2x}dx=\int\dfrac{dx}{cos^2x}+\int\dfrac{-sinx.dx}{cos^2x}=\int\dfrac{dx}{cos^2x}+\int\dfrac{d\left(cosx\right)}{cos^2x}\)

\(=tanx-\dfrac{1}{cosx}+C\)

\(I=\int\limits^2_1\dfrac{\left(x+1\right)^2-2}{x+1}dx=\int\limits^2_1\left(x+1-\dfrac{2}{x+1}\right)dx\)

\(=\left(\dfrac{x^2}{2}+x-2ln\left|x+1\right|\right)|^2_1=\dfrac{5}{2}-2ln\dfrac{3}{2}\)

\(I=\int\left(tanx+tan^3x\right)dx=\int tanx\left(1+tan^2x\right)dx\)

\(I=\int tanx\dfrac{1}{cos^2x}dx\)

Đặt \(tanx=t\Rightarrow\dfrac{dx}{cos^2x}=dt\)

\(\Rightarrow I=\int t.dt=\dfrac{t^2}{2}+C=\dfrac{tan^2x}{2}+C\)

\(I=\int\dfrac{lnx}{x\sqrt{1+lnx}}dx\)

Đặt \(\sqrt{1+lnx}=t\Rightarrow lnx=t^2-1\Rightarrow\dfrac{dx}{x}=2t.dt\)

\(\Rightarrow I=\int\dfrac{\left(t^2-1\right)}{t}.2tdt=\int\left(2t^2-2\right)dt=\dfrac{2t^3}{3}-2t+C\)

\(\Rightarrow I=\dfrac{2\left(1+lnx\right)\sqrt{1+lnx}}{3}-2\sqrt{1+lnx}+C\)

\(I=\int\dfrac{x+1}{\sqrt{x+1}}dx=\int2\left(\sqrt{x+1}\right)^2.\dfrac{dx}{2\sqrt{x+1}}=2\int\left(\sqrt{x+1}\right)^2.d\left(\sqrt{x+1}\right)\)

\(=\dfrac{2}{3}\sqrt{\left(x+1\right)^3}+C=\dfrac{2}{3}\left(x+1\right)\sqrt{x+1}+C\)

\(\int\dfrac{dx}{\left(\sqrt{2}.sin\left(x-\dfrac{\pi}{4}\right)\right)^2}=\dfrac{1}{2}\int\dfrac{dx}{sin^2\left(x-\dfrac{\pi}{4}\right)}=\dfrac{-cot\left(x-\dfrac{\pi}{4}\right)}{2}+C\)

\(\int sin^4xdx=\int\left(\dfrac{1-cos2x}{2}\right)^2dx=\dfrac{1}{4}\int\left(cos^22x-2cos2x+1\right)dx\)

\(=\dfrac{1}{4}\int\left(\dfrac{1}{2}+\dfrac{1}{2}cos4x-2cos2x+1\right)dx=\int\left(\dfrac{3}{8}+\dfrac{cos4x}{8}-\dfrac{cos2x}{2}\right)dx\)

\(=\dfrac{3x}{8}+\dfrac{sin4x}{32}-\dfrac{sin2x}{4}+C\)

\(\int\dfrac{1}{xln^5x}dx=\int ln^{-5}x.d\left(lnx\right)=\dfrac{ln^{-4}x}{-4}+C=\dfrac{-1}{4ln^4x}+C\)

Lời giải:
\(\int \frac{4x^2+3x+1}{1-2x}dx=-\int \frac{4x^2+3x+1}{2x-1}dx=-\frac{1}{2}\int \frac{8x^2+6x+2}{2x-1}dx\)

\(=-\frac{1}{2}\int \frac{4x(2x-1)+5(2x-1)+7}{2x-1}dx\)

\(=-\frac{1}{2}\int (4x+5+\frac{7}{2x-1})dx\)

\(=-\frac{1}{2}\int (4x+5)dx-\frac{1}{2}\int \frac{7}{2x-1}dx\)

\(=-\frac{1}{2}\int (4x+5)dx-\frac{7}{4}\int \frac{d(2x-1)}{2x-1}\)

\(=-\frac{1}{2}(2x^2+5x)-\frac{7}{4}\ln |2x-1|+c\)

\(f\left(x\right)=\dfrac{2^{x+1}}{10^x}-\dfrac{5^{x-1}}{10^x}=\dfrac{2}{5^x}-\dfrac{1}{5.2^x}=2.5^{-x}-\dfrac{1}{5}.2^{-x}\)

\(\Rightarrow\int f\left(x\right)dx=\int\left(2.5^{-x}-\dfrac{1}{5}.2^{-x}\right)dx=\int\left(-2.5^{-x}+\dfrac{1}{5}.2^{-x}\right)d\left(-x\right)\)

\(=\dfrac{-2.5^{-x}}{ln5}+\dfrac{2^{-x}}{5.ln2}+C=\dfrac{1}{2^x.5ln2}-\dfrac{2}{5^x.ln5}+C\)

\(\int cosx.sin^2x.dx=\int sin^2x.d\left(sinx\right)=\dfrac{sin^3x}{3}+C\)