Nhìn đề dữ dội y hệt cr của tui z :( Để làm từ từ 

Lập bảng xét dấu cho \(\left|x^2-1\right|\) trên đoạn \(\left[-2;2\right]\)

\(\left(-2;-1\right):+\)

\(\left(-1;1\right):-\)

\(\left(1;2\right):+\)

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

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

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

Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính 

2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)

\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)

\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)

3/ \(I=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx+\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)

Xét \(A=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx\)

\(t=\sin x\Rightarrow dt=\cos x.dx\Rightarrow A=\int\limits^{\dfrac{\pi}{2}}_0e^t.dt=e^{\sin x}|^{\dfrac{\pi}{2}}_0\)

Xét \(B=\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)

\(=\int\limits^{\dfrac{\pi}{2}}_0\dfrac{1+\cos2x}{2}.dx=\dfrac{1}{2}.\int\limits^{\dfrac{\pi}{2}}_0dx+\dfrac{1}{2}\int\limits^{\dfrac{\pi}{2}}_0\cos2x.dx\)

\(=\dfrac{1}{2}x|^{\dfrac{\pi}{2}}_0+\dfrac{1}{2}.\dfrac{1}{2}\sin2x|^{\dfrac{\pi}{2}}_0\)

I=A+B=...

a.

Đặt \(\sqrt{1-x^2}=u\Rightarrow x^2=1-u^2\Rightarrow xdx=-udu\)

\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=0\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^0_1\left(1-u^2\right).u.\left(-udu\right)=\int\limits^1_0\left(u^2-u^4\right)du=\left(\dfrac{1}{3}u^3-\dfrac{1}{5}u^5\right)|^1_0\)

\(=\dfrac{2}{15}\)

b.

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

Đặt \(x-1=tanu\Rightarrow dx=\dfrac{1}{cos^2u}du\)

\(\left\{{}\begin{matrix}x=1\Rightarrow u=0\\x=2\Rightarrow u=\dfrac{\pi}{4}\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{1}{tan^2u+1}.\dfrac{1}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{cos^2u}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0du\)

\(=u|^{\dfrac{\pi}{4}}_0=\dfrac{\pi}{4}\)

c.

\(\int\limits^2_1\dfrac{dx}{\sqrt{4-x^2}}\)

Đặt \(x=2sinu\Rightarrow dx=2cosu.du\)

\(\left\{{}\begin{matrix}x=1\Rightarrow u=\dfrac{\pi}{6}\\x=2\Rightarrow u=\dfrac{\pi}{2}\end{matrix}\right.\)

\(I=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{2cosu.du}{\sqrt{4-4sin^2u}}=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{2cosu.du}{2cosu}=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}du\)

\(=u|^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}=\dfrac{\pi}{3}\)

a.

\(\int\limits^{\sqrt{7}}_0\dfrac{x^3}{\sqrt[3]{x^2+1}}dx\)

Đặt \(\sqrt[3]{x^2+1}=u\Rightarrow x^2+1=u^3\Rightarrow x^2=u^3-1\Rightarrow x.dx=\dfrac{3}{2}u^2du\)

\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=\sqrt{7}\Rightarrow u=2\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^2_1\dfrac{\left(u^3-1\right).\dfrac{3}{2}u^2du}{u}=\int\limits^2_1\dfrac{3}{2}\left(u^4-u\right)du=\dfrac{3}{2}\left(\dfrac{1}{5}u^5-\dfrac{1}{2}u^2\right)|^2_1\)

\(=\dfrac{141}{20}\)

b.

Đặt \(\sqrt{x+3}=u\Rightarrow x=u^2-3\Rightarrow dx=2udu\)

\(\left\{{}\begin{matrix}x=1\Rightarrow u=2\\x=6\Rightarrow u=3\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^3_2\dfrac{u+1}{u^2-3+2}.2udu=\int\limits^3_2\dfrac{2udu}{u-1}=\int\limits^3_22\left(1+\dfrac{1}{u-1}\right)du\)

\(=2\left(u+ln\left|u-1\right|\right)|^3_2=2\left(1+ln2\right)\)

Lời giải:
\(=\int ^1_0\frac{(2x-7)(x^2+2x+1)+13(x+1)-10}{x^2+2x+1}dx=\int ^1_0(2x-7)dx+\int ^1_0\frac{13}{x+1}dx-\int ^1_0\frac{10dx}{(x+1)^2}\)

\(=|^1_0(x^2-7x)+13|^1_0\ln |x+1|+|^1_0\frac{10}{x+1}\)

\(=-11+13\ln 2\)

\(2I=\int\limits^4_0\left(e^x\sqrt{2x+1}+\dfrac{e^x}{\sqrt{2x+1}}\right)dx=\int\limits^4_0e^x\sqrt{2x+1}dx+\int\limits^4_0\dfrac{e^x}{\sqrt{2x+1}}dx=I_1+I_2\)

Xét \(I_1=\int\limits^4_0e^x\sqrt{2x+1}dx\)

Đặt \(\left\{{}\begin{matrix}u=\sqrt{2x+1}\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{1}{\sqrt{2x+1}}dx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I_1=e^x.\sqrt{2x+1}|^4_0-\int\limits^4_0\dfrac{e^x}{\sqrt{2x+1}}dx=3e^4-1-I_2\)

Do đó:

\(2I=3e^4-1-I_2+I_2=3e^4-1\)

\(\Rightarrow I=\dfrac{3}{2}e^4-\dfrac{1}{2}\Rightarrow a=\dfrac{3}{2};b=4;c=-\dfrac{1}{2}\)

Câu 1:

Ta có \(I_1=\int ^{1}_{0}\frac{4x+2}{x^2+x+1}dx=2\int ^{1}_{0}\frac{2x+1}{x^2+x+1}dx\)

\(=2\int ^{1}_{0}\frac{d(x^2+x+1)}{x^2+x+1}=2.\left.\begin{matrix} 1\\ 0\end{matrix}\right|\ln |x^2+x+1|=2\ln 3\)

Câu 2:

\(I_2=\int ^{1}_{0}\frac{4x+1}{(2-x)^4}dx=\int ^{1}_{0}\frac{4(x-2)+9}{(2-x)^4}dx\)

\(=4\int ^{1}_{0}\frac{dx}{(x-2)^3}+9\int \frac{dx}{(2-x)^4}=4\int ^{1}_{0}\frac{d(x-2)}{(x-2)^3}-9\int ^{1}_{0}\frac{d(2-x)}{(2-x)^4}\)

\(=4\int ^{-1}_{-2}\frac{dt}{t^3}-9\int ^{1}_{2}\frac{dk}{k^4}\) với \(x-2=t; 2-x=k\)

\(=4.\left.\begin{matrix} -1\\ -2\end{matrix}\right|\frac{t^{-3+1}}{-3+1}-9.\left.\begin{matrix} 1\\ 2\end{matrix}\right|\frac{k^{-4+1}}{-4+1}=\frac{9}{8}\)

Câu 3:

Phân số \(\frac{x^2+1}{(x^3+3x)^3}\) không xác định trên \([0;1]\); hàm không liên tục nên không có tích phân.