Chương 3: NGUYÊN HÀM. TÍCH PHÂN VÀ ỨNG DỤNG

Tìm đạo hàm của hàm số nào vậy nhỉ bạn?

Đặt \(\sqrt{x}=t\Rightarrow x=t^2\Rightarrow dx=2tdt\)

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

Lời giải:
\(I=\int ^1_{-1}\frac{1}{(2-x)(2+x)}dx=\frac{1}{4}\int ^1_{-1}(\frac{1}{2-x}+\frac{1}{2+x})dx\)

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

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

\(=(-\frac{1}{4}\ln |x-2|+\frac{1}{4}\ln |x+2|)^1_{-1}=\frac{1}{2}\ln 3\)

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

⇒  \(\int\left(x+2\right)\ln xdx=\ln x\left(\dfrac{1}{2}x^2+2x\right)-\int\left(\dfrac{1}{2}x^2+2x\right)\dfrac{1}{x}dx\)

\(=\ln x\left(\dfrac{1}{2}x^2+2x\right)-\dfrac{1}{6}x^3-x^2+C\)

\(=\int\left(1+x\right)^ndx=\dfrac{1}{n+1}\left(1+x\right)^{n+1}+C\)

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

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

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

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

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

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

\(\Rightarrow I=\int\limits^0_1\left(1-u^2\right).u.\left(-2udu\right)=\int\limits^1_0\left(2u^2-2u^4\right)du=\left(\dfrac{2}{3}u^3-\dfrac{2}{5}u^5\right)|^1_0=\dfrac{4}{15}\)

a. \(\int\dfrac{x^3}{x-2}dx=\int\left(x^2+2x+4+\dfrac{8}{x-2}\right)dx=\dfrac{1}{3}x^3+x^2+4x+8ln\left|x-2\right|+C\)

b. \(\int\dfrac{dx}{x\sqrt{x^2+1}}=\int\dfrac{xdx}{x^2\sqrt{x^2+1}}\)

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

\(I=\int\dfrac{udu}{\left(u^2-1\right)u}=\int\dfrac{du}{u^2-1}=\dfrac{1}{2}\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right)du=\dfrac{1}{2}ln\left|\dfrac{u-1}{u+1}\right|+C\)

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

c. \(\int\left(\dfrac{5}{x}+\sqrt{x^3}\right)dx=\int\left(\dfrac{5}{x}+x^{\dfrac{3}{2}}\right)dx=5ln\left|x\right|+\dfrac{2}{5}\sqrt{x^5}+C\)

d. \(\int\dfrac{x\sqrt{x}+\sqrt{x}}{x^2}dx=\int\left(x^{-\dfrac{1}{2}}+x^{-\dfrac{3}{2}}\right)dx=2\sqrt{x}-\dfrac{1}{2\sqrt{x}}+C\)

e. \(\int\dfrac{dx}{\sqrt{1-x^2}}=arcsin\left(x\right)+C\)