Lời giải:
\(=\int ^1_0\frac{(2x-7)(x^2+2x+1)+13(x+1)-10}{x^2+2x+1}dx=\int ^1_0(2x-7)dx+\int ^1_0\frac{13}{x+1}dx-\int ^1_0\frac{10dx}{(x+1)^2}\)
\(=|^1_0(x^2-7x)+13|^1_0\ln |x+1|+|^1_0\frac{10}{x+1}\)
\(=-11+13\ln 2\)
Lời giải:
\(=\int ^1_0\frac{(2x-7)(x^2+2x+1)+13(x+1)-10}{x^2+2x+1}dx=\int ^1_0(2x-7)dx+\int ^1_0\frac{13}{x+1}dx-\int ^1_0\frac{10dx}{(x+1)^2}\)
\(=|^1_0(x^2-7x)+13|^1_0\ln |x+1|+|^1_0\frac{10}{x+1}\)
\(=-11+13\ln 2\)
Tính tích phân sau: \(\int_0^1\frac{x^4-2x^3-4x^2+x-2}{x^2-2x-3}dx\)
a\(\int_0^1\dfrac{dx}{x^4+4x^2+3}\)
b \(\int\dfrac{x^2-1}{x^4+1}\)
c\(\int\dfrac{dx}{x\left(x^3+1\right)}\)
d \(\int_0^1\dfrac{xdx}{x^4+x^2+1}\)
\(\int_0^{\sqrt{7}}\dfrac{x^3}{\sqrt[3]{x^2+1}}dx\)
\(\int_1^6\dfrac{\sqrt{x+3}+1}{x+2}dx\)
Câu 1. Cho hàm số chẵn y=f (x) liên tục trên R và \(\int\limits^1_{-1}\dfrac{f\left(2x\right)}{1+2^x}dx=8\).Tính \(\int_0^2f\left(x\right)dx\)
Câu 2:Cho hàm số y=f (x) có đạo hàm và liên tục trên [0;1]và thỏa f(0)=1.\(\int_0^1\left[f'\left(x\right)\left[f^2\left(x\right)\right]+1\right]dx=2\int_0^1\sqrt{f'\left(x\right)}f\left(x\right)dx\).Tính\(\int_0^1\left[f^3\left(x\right)\right]dx\).
\(\int_0^{\dfrac{\pi}{4}}\) \(\dfrac{3sin^2x-4cos^2x}{cos^2x}dx\)
a) \(\int_{\dfrac{\pi}{8}}^{\dfrac{2\pi}{8}}\)\(\dfrac{dx}{sin^2xcos^2x}\)
b) \(\int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}\)\(\dfrac{cos2xdx}{sin^2xcos^2x}\)
c) \(\int_0^{\dfrac{\pi}{3}}\)\(\dfrac{cos3x}{cosx}\)dx
\(\int_0^{\dfrac{\pi}{4}}\)\(\dfrac{cos2x-3sin^2x}{cos^2x}dx\)
cho \(\int_0^1\frac{x^3+2x^2+3}{x+2}dx=\frac{1}{a}+bln\frac{3}{2}\left(a,b>0\right)TínhS=a^2+b^2\)
\(\int_0^1\)\(\dfrac{x-2}{\left(x+1\right)^2}dx\)