Bài 3a. Tính nguyên hàm - tích phân bằng phương pháp đổi biến số

\(\int\frac{x-3}{\left(x-1\right)\left(x+3\right)}dx=\int\left(\frac{3}{2\left(x+3\right)}-\frac{1}{2\left(x-1\right)}\right)dx=\frac{3}{2}ln\left|x+3\right|-\frac{1}{2}ln\left|x-1\right|+C\)

\(\int\frac{x}{x^2-4}dx=\frac{1}{2}\int\frac{d\left(x^2-4\right)}{x^2-4}=\frac{1}{2}ln\left|x^2-4\right|+C\)

Bài 1:

\(F'\left(x\right)=e^x+\left(x-1\right)e^x=xe^x=\frac{x}{e^x}.e^{2x}\Rightarrow f\left(x\right)=\frac{x}{e^x}\)

Xét \(I=\int f'\left(x\right)e^{2x}dx\)

Đặt \(\left\{{}\begin{matrix}u=e^{2x}\\v=f'\left(x\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2e^{2x}dx\\v=f\left(x\right)\end{matrix}\right.\)

\(\Rightarrow I=f\left(x\right).e^{2x}+2\int f\left(x\right).e^{2x}dx=x.e^x+2\left(x-1\right)e^x+C=\left(3x-2\right)e^x+C\)

2.

Xét \(J=\int\limits^1_0xf\left(6x\right)dx\)

Đặt \(6x=t\Rightarrow dx=\frac{1}{6}dt\Rightarrow J=\frac{1}{36}\int\limits^6_0t.f\left(t\right)dt=\frac{1}{36}\int\limits^6_0x.f\left(x\right)dx=1\)

\(\Rightarrow I=\int\limits^6_0x.f\left(x\right)dx=36\)

Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\frac{1}{2}x^2\end{matrix}\right.\)

\(\Rightarrow I=\frac{1}{2}x^2f\left(x\right)|^6_0-\frac{1}{2}\int\limits^6_0x^2.f'\left(x\right)dx\)

\(\Leftrightarrow36=18-\frac{1}{2}\int\limits^6_0x^2f'\left(x\right)dx\)

\(\Rightarrow\int\limits^6_0x^2f'\left(x\right)dx=-36\)

Lấy tích phân 2 vế:

\(\int\limits^1_0\left[f'\left(x\right)\right]^2dx+\int\limits^1_04\left(6x^2-1\right)f\left(x\right)dx=\int\limits^1_0\left(40x^6-44x^4+32x^2-4\right)dx=\frac{376}{105}\)

Xét \(I=\int\limits^1_0\left(6x^2-1\right)f\left(x\right)dx\)

Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=\left(6x^2-1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=2x^3-x\end{matrix}\right.\)

\(\Rightarrow I=\left(2x^3-x\right)f\left(x\right)|^1_0-\int\limits^1_0\left(2x^3-x\right)f'\left(x\right)dx=1-\int\limits^1_0\left(2x^3-x\right)f'\left(x\right)dx\)

\(\Rightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx+4-4\int\limits^1_0\left(2x^3-x\right)f'\left(x\right)dx=\frac{376}{105}\)

\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)-2\left(2x^3-x\right)\right]^2dx-\int\limits^1_04\left(2x^3-x\right)^2dx=-\frac{44}{105}\)

\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)-2\left(2x^3-x\right)\right]^2dx-\frac{44}{105}=-\frac{44}{105}\)

\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)-\left(4x^3-2x\right)\right]^2dx=0\)

\(\Rightarrow f'\left(x\right)=4x^3-2x\Rightarrow f\left(x\right)=x^4-x^2+C\)

\(f\left(1\right)=1\Rightarrow1-1+C=1\Rightarrow C=1\)

\(\Rightarrow f\left(x\right)=x^4-x^2+1\)

\(\Rightarrow\int\limits^1_0x\left(x^4-x^2+1\right)dx=\frac{5}{12}\)

\(\int-\frac{5}{16\left(1-x\right)}+\frac{7}{4\left(1-x\right)^2}-\frac{5}{16\left(x+3\right)}dx\)

= \(-\int\frac{5}{16\left(1-x\right)}dx+\int\frac{7}{4\left(1-x\right)^2}dc-\int\frac{5}{16\left(x+3\right)}dx\)

= \(\frac{5}{16}ln\left|1-x\right|+\frac{7}{4-4x}-\frac{5}{16}ln\left|x+3\right|\)+C

Đặt \(\left\{{}\begin{matrix}u=ln\left(x+1\right)\\dv=\left(x^2-1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{dx}{x+1}\\v=\frac{1}{3}x^3-x\end{matrix}\right.\)

\(\Rightarrow I=\left(\frac{1}{3}x^3-x\right)ln\left(x+1\right)-\frac{1}{3}\int\frac{\left(x^3-3x\right)}{x+1}dx\)

Xét \(J=\int\frac{\left(x^3-3x\right)dx}{x+1}=\int\left(x^2-x-2+\frac{2}{x+1}\right)dx\)

\(\Rightarrow J=\frac{1}{3}x^3-\frac{1}{2}x^2-2x+2ln\left(x+1\right)\)

\(\Rightarrow I=\left(\frac{1}{3}x^3-x\right)ln\left(x+1\right)-\frac{1}{3}\left(\frac{1}{3}x^3-\frac{1}{2}x^2-2x+2ln\left(x+1\right)\right)+C\)

Bạn tự rút gọn biểu thức cuối

\(I=\int e^xcosxdx\)

Đặt \(\left\{{}\begin{matrix}u=e^x\\dv=cosxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=e^xdx\\v=sinx\end{matrix}\right.\)

\(\Rightarrow I=e^xsinx-\int e^xsinxdx\)

Xét \(J=\int e^xsinxdx\Rightarrow\left\{{}\begin{matrix}u=e^x\\dv=sinxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=e^x\\v=-cosx\end{matrix}\right.\)

\(\Rightarrow J=-e^xcosx+\int e^xcosxdx=-e^xcosx+C\)

\(\Rightarrow I=e^xsinx-\left(-e^xcosx+I\right)=e^x\left(sinx+cosx\right)-I\)

\(\Rightarrow2I=e^x\left(sinx+cosx\right)\Rightarrow I=\left(\frac{1}{2}cosx+\frac{1}{2}sinx\right)e^x\)

Hoặc đơn giản là đạo hàm F(x) và đồng nhất hệ số với f(x) là xong

\(I=\int\limits^1_0\frac{xdx}{\sqrt{3x+1}+\sqrt{2x+1}}=\int\limits^1_0\frac{x\left(\sqrt{3x+1}-\sqrt{2x+1}\right)}{x}dx\)

\(=\int\limits^1_0\left(\left(3x+1\right)^{\frac{1}{2}}-\left(2x+1\right)^{\frac{1}{2}}\right)dx=\left[\frac{2}{9}\left(3x+1\right)^{\frac{3}{2}}-\frac{1}{3}\left(2x+1\right)^{\frac{3}{2}}\right]|^1_0\)

\(=\frac{2}{9}\sqrt{4^3}-\frac{1}{3}\sqrt{3^3}-\frac{2}{9}+\frac{1}{3}=\frac{17-9\sqrt{3}}{9}\)

\(\Rightarrow a+b=17-9=8\)