Bài tập 5 trang 113 SGK Giải tích 12

Câu a:

Đặt u = 1 + 3x ta có: du = 3dx

Khi x = 0 thì u = 1; khi x = 1 thì u = 4.

Do đó: \(\int_{0}^{1}(1+3x)^{\frac{3}{2}}dx =\frac{1}{3}\int_{0}^{4}u^{\frac{3}{2}}du\)

 \(=\frac{1}{3}\frac{2}{5}u^{\frac{5}{2}} \Bigg|_{0}^{4}= \frac{2}{15}(4^{\frac{5}{2}}-1)=\frac{62}{15}\)

Câu b:

  \(= \int_{0}^{\frac{1}{2}}\frac{x^2+x+1}{x+1}dx=\int_{0}^{\frac{1}{2}}(x+\frac{1}{x+1})dx\)    

 \(=\frac{x^{2}}{2}\Bigg|_{0}^{\frac{1}{2}}+ln\left | x+1 \right | \Bigg|_{0}^{\frac{1}{2}}=\frac{1}{8}+ln\frac{3}{2}\)

Câu c:

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

Ta có:

\(\int_{1}^{2}\frac{ln(1+x)}{x^{2}}dx =-\frac{1}{x}ln(1+x) \Bigg|^2_1+ \int_{1}^{2} \frac{dx}{x(1+x)}\)

\(=-\frac{1}{2}ln 3+ln 2 +\int_{1}^{2}\frac{dx}{x(1+x)}\)

Xét \(\int_{1}^{2}\frac{dx}{(x+1)x}\). Ta có: \(\frac{1}{x(1+x)}=\frac{1}{x}-\frac{1}{x+1}\)

Do đó:

\(\int_{1}^{2}\frac{dx}{x(x+1)}=\int_{1}^{2}\left ( \frac{1}{x}- \frac{1}{x+1} \right )dx=\int_{1}^{2}\frac{dx}{x}-\int_{1}^{2}\frac{dx}{x+1}\)

\(=lnx \Bigg|^2_1-ln(1+x)\Bigg|^2_1=ln2 -ln3+ln2\)

Vậy \(\int_{1}^{2}\frac{ln(1+x)}{x^2}dx=-\frac{1}{2}ln3 +ln2 +ln2-ln3+ln2\)

\(=3ln2-\frac{3}{2}ln3=ln8-\frac{3}{2}ln3=3ln \frac{2\sqrt{3}}{3}\)

-- Mod Toán 12 HỌC247

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