Một nguyên hàm của hàm số f(x) = x^2 + 1/ x^4 + 2x^3 - 10x^2 - 2x + 1 có dạng F(x) = a/b ln trị tuyệt đối x^2 - cx - 1 / x^2 + dx - 1

A. 24

B. 21

C. 15

D. 13

Trả lời

Chọn D

$\int \frac{x^2+1}{x^4+2x^3-10x^2-2x+1}\text{d}x=\int \frac{1+\frac{1}{x^2}}{x^2+2x-10-2\frac{1}{x}+\frac{1}{x^2}}\text{d}x=\int \frac{\text{d}\left(x-\frac{1}{x}\right)}{{\left(x-\frac{1}{x}\right)}^2+2\left(x-\frac{1}{x}\right)-8}$

$=\frac{1}{6}\int \frac{\left(x-\frac{1}{x}+4\right)-\left(x-\frac{1}{x}-2\right)}{\left(x-\frac{1}{x}-2\right)\left(x-\frac{1}{x}+4\right)}\text{d}\left(x-\frac{1}{x}\right)=\frac{1}{6}\int \left[\frac{\text{d}\left(x-\frac{1}{x}\right)}{x-\frac{1}{x}-2}-\frac{\text{d}\left(x-\frac{1}{x}\right)}{x-\frac{1}{x}+4}\right]$

$=\frac{1}{6}\ln \left|x-\frac{1}{x}-2\right|-\frac{1}{6}\ln \left|x-\frac{1}{x}+4\right|+C=\frac{1}{6}\ln \left|\frac{x-\frac{1}{x}-2}{x-\frac{1}{x}+4}\right|+C=\frac{1}{6}\ln \left|\frac{x^2-2x-1}{x^2+4x-1}\right|+C$