Question

[Nguyễn Hữu Thắng]

Giúp em với : Tính nguyên hàm sau

Answer 1

[Akai Haruma]

Lời giải:

Phân tích:

\(I=\int\frac{x^8}{x^3+x+2}dx=\underbrace{\int (x^5-x^3-2x^2+x+4)dx}_{A}+\underbrace{\int \frac{3x^2-6x-8}{x^3+x+2}dx}_{B}\)

Có \(A=\frac{x^6}{6}-\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+4x+c(1)\)

\(B=\int\frac{3(x^2-x+2)-3(x+1)-11}{(x^2-x+2)(x+1)}dx\) \(=3\int\frac{dx}{x+1}-3\int\frac{dx}{x^2-x+2}-\int\frac{11dx}{x^3+x+2}\)

Đối với \(\int\frac{dx}{x^2-x+2}=\int\frac{dx}{(x-\frac{1}{2})^2+\frac{7}{4}}\) ta đặt \(x-\frac{1}{2}=\frac{\sqrt{7}}{2}\tan t\)

\(\Rightarrow \int\frac{dx}{x^2-x+2}=\frac{2\sqrt{7}}{7}\tan ^-1\left(\frac{2x-1}{\sqrt{7}}\right)+c\)

Đối với

\(\int\frac{dx}{x^3+x+2}=\int\frac{d(x^3+x+2)}{x^3+x+2}-\int\frac{3x^2dx}{x^3+x+2}=\ln|x^3+x+2|-\int\frac{3dx}{x+1}-\int\frac{3dx}{x^2-x+2}+\int\frac{9dx}{x^3+x+2}\)

\(\Rightarrow -8\int\frac{dx}{x^3+x+2}=\ln|x^3+x+2|-3\ln|x+1|-\frac{6\sqrt{7}}{7}\tan^{-1}\left(\frac{2x-1}{\sqrt{7}}\right)\)

\(\Rightarrow \int\frac{dx}{x^3+x+2}=\frac{-\ln|x^2-x+2|}{8}+\frac{\ln|x+1|}{4}+\frac{3\sqrt{7}}{28}\tan^-1\left(\frac{2x-1}{\sqrt{7}}\right)\)

Vậy \(B=\frac{\ln|x+1|}{4}+\frac{11\ln|x^2-x+2|}{8}-\frac{57\sqrt{7}}{28}\tan^-1\left(\frac{2x-1}{\sqrt{7}}\right)+c(2)\)

Từ \((1),(2)\Rightarrow I=\frac{x^6}{6}-\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+4x+\frac{\ln|x+1|}{4}+\frac{11\ln|x^2-x+2|}{8}-\frac{57\sqrt{7}}{28}\tan^-1\left(\frac{2x-1}{\sqrt{7}}\right)+c\)

Answer 2

[Nguyễn Trọng Đức]

Lớp 12 là lớp lớn nhất rồi a còn xưng em em làm j???