Bài tập Nguyên hàm hữu tỉ

\(\mathrm{I}=\int \frac{\mathrm{P}(\mathrm{x}) \mathrm{dx}}{\mathrm{ax}+\mathrm{b}}\)

phân tích:

\(\frac{P(x)}{a x+b}=g(x)+\frac{k}{a x+b}\)

khi đó

\(I=\int g(x) d x+k \int \frac{d x}{a x+b}\)

\(I=\int \frac{m x+n}{a x^{2}+b x+c} d x\)

Trường hợp 1:

\(\Delta=b^{2}-4 a c\gt 0\)

phân tích:

\(\frac{m x+n}{a x^{2}+b x+c}=\frac{m x+m}{a\left(x-x_{1}\right)\left(x-x_{2}\right)}=\frac{1}{a}\left(\frac{A}{x-x_{1}}+\frac{B}{x-x_{2}}\right)\)

(đồng nhất hệ số để tìm A, B)

=> \(I=\frac{1}{a}\left(A \ln \left|x-x_{1}\right|+B \ln \left|x-x_{2}\right|\right)+C\).

Trường hợp 2:

\(\Delta=b^{2}-4 a c=0\)

\(\frac{m x+n}{a \cdot x^{2}+b x+c}=\frac{m x+n}{a\left(x-x_{0}\right)^{2}}=\frac{m\left(x-x_{0}\right)+p}{a\left(x-x_{0}\right)^{2}}\)

= \(\frac{\mathrm{m}}{\mathrm{a}\left(\mathrm{x}-\mathrm{x}_{0}\right)}+\frac{\mathrm{P}}{\mathrm{a}\left(\mathrm{x}-\mathrm{x}_{0}\right)^{2}}\)

Trường hợp 3:

\(\Delta=\mathrm{b}^{2}-4 \mathrm{ac}\lt 0\)

phân tích:

\(\frac{m x+n}{a x^{2}+b x+c}=\frac{k(2 a x+b)}{a x^{2}+b x+c}+\frac{p}{a\left(x-x_{0}\right)^{2}+q}\)

khi đó:

\(I=\int \frac{k d\left(a x^{2}+b x+c\right)}{a x^{2}+b x+c}+\frac{p}{a} \int \frac{1}{\left(x-x_{0}\right)^{2}+n^{2}} d x\)

\(\mathrm{I}=\int \frac{\mathrm{P}(\mathrm{x}) \mathrm{dx}}{\mathrm{Q}(\mathrm{x})}\) với \(\mathrm{Q}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}\)

Trường hợp 1

\(a x^{3}+b x^{2}+c x+d=a\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\)

Phân tích

\(\frac{P(x)}{a x^{3}+b x^{2}+c x+d}=\frac{A}{x-x_{1}}+\frac{B}{x-x_{2}}+\frac{C}{x-x_{3}}\)

Trường hợp 2

\(a x^{3}+x^{2}+c x+d=a\left(x-x_{1}\right)\left(x-x_{2}\right)^{2}\)

phân tích

\(\frac{P(x)}{a x^{3}+b x^{2}+c x+d}=\frac{A}{x-x_{1}}+\frac{B x+C}{\left(x-x_{2}\right)^{2}}\)

Trường hợp 3

\(\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}=\mathrm{a}\left(\mathrm{x}-\mathrm{x}_{1}\right)\left(\mathrm{mx}^{2}+\mathrm{nx}+\mathrm{p}\right)\) 

trong đó \(\mathrm{mx}^{2}+\mathrm{nx}+\mathrm{p}=0\) vô nghiệm

phân tích:

\(\frac{\mathrm{P}(\mathrm{x})}{\mathrm{ax}+\mathrm{bx}+\mathrm{cx}+\mathrm{d}}=\frac{\mathrm{A}}{\mathrm{x}-\mathrm{x}_{1}^{2}}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{m} \mathrm{x}^{2}+\mathrm{nx}+\mathrm{p}}\)

\(\mathrm{I}=\int \frac{\mathrm{P}(\mathrm{x}) \mathrm{dx}}{\mathrm{x}^{4} \pm \mathrm{a}^{2}}\) trong đó bậc của \(\mathrm{P}(\mathrm{x})\) nhỏ hơn 4

Trường hợp 1: \(I=\int \frac{P(x) d x}{x^{4}+a^{2}}\)

phân tích: 

\(\frac{P(x)}{x^{4}+a^{2}}=\frac{A\left(x^{2}+a\right)+B\left(x^{2}-a\right)+C x^{3}+D x}{x^{4}+a^{2}}\)

khi đó ta có 

\(I_{1}=\int \frac{x^{2}+a}{x^{4}+a^{2}} d x=\int \frac{1+\frac{a}{x^{2}}}{x^{2}+\frac{a^{2}}{x^{2}}} d x\)

\(\int \frac{d\left(x-\frac{a}{x}\right)}{\left(x-\frac{a}{x}\right)^{2}+2 a} \rightarrow I_{1}=\int \frac{d u}{u^{2}+2 a}\)

\(I_{2}=\int \frac{x^{2}-a}{x^{4}+a^{2}} d x=\int \frac{1-\frac{a}{x^{2}}}{x^{2}+\frac{a^{2}}{x^{2}}} d x=\)

\(\int \frac{d\left(x+\frac{a}{x}\right)}{\left(x+\frac{a}{x}\right)^{2}-2 a} \rightarrow I_{2}=\int \frac{d u}{u^{2}-2 a}\)

\(I_{3}=\int \frac{x^{3} d x}{x^{4}+a^{2}}=\frac{1}{4} \int \frac{d\left(x^{4}+a^{2}\right)}{x^{4}+a^{2}}\)

=\(\frac{1}{4} \ln \left|x^{4}+a^{2}\right|+C\)

\(I_{4}=\int \frac{x d x}{x^{4}+a^{2}}=\frac{1}{2} \int \frac{d\left(x^{2}\right)}{x^{4}+a^{2}}\)

=> \(I_{4}=\frac{1}{2} \int \frac{d u}{u^{2}+a^{2}}\)

từ đó ta suy ra nguyên hàm   \(I=\int \frac{P(x) d x}{x^{4}+a^{2}}\)

Một số nguyên hàm hữu tỷ khi Q(x) là đa thức bậc 6

- \(\mathrm{I}_{1}=\int \frac{\mathrm{dx}}{\mathrm{x}^{6}-1}=\int \frac{\mathrm{dx}}{\left(\mathrm{x}^{3}-1\right)\left(\mathrm{x}^{3}+1\right)}=\frac{1}{2} \int\left(\frac{1}{\mathrm{x}^{3}-1}-\frac{1}{\mathrm{x}^{3}+1}\right)\)

- \(\mathrm{I}_{2}=\int \frac{\mathrm{xdx}}{\mathrm{x}^{6}-1}=\frac{1}{2} \int \frac{\mathrm{dx}^{2}}{\left(\mathrm{x}^{2}\right)^{3}-1} \rightarrow \mathrm{I}_{2}=\frac{1}{2} \int \frac{\mathrm{du}}{\mathrm{u}^{3}-1}\)

- \(I_{3}=\int \frac{x^{2} d x}{x^{6}-1}=\frac{1}{3} \int \frac{d\left(x^{3}\right)}{x^{6}-1} \rightarrow I_{3}=\frac{1}{3} \int \frac{d u}{u^{2}-1}\)

- \(I_{4}=\int \frac{x^{3} d x}{x^{6}-1}=\frac{1}{2} \int \frac{x^{2} d\left(x^{2}\right)}{x^{6}-1} \rightarrow I_{4}=\frac{1}{2} \int \frac{u d u}{u^{3}-1}\)

- \(I_{5}=\int \frac{x^{4} d x}{x^{6}-1}=\int \frac{\left(x^{4}+x^{2}+1\right)+\left(x^{2}-1\right)-2}{\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)} d x\)

= \(\int \frac{d x}{x^{2}-1}-\int \frac{d x}{x^{4}+x^{2}+1}-2 \int \frac{d x}{x^{6}-1}\)

với K = \(\int \frac{\mathrm{dx}}{\mathrm{x}^{4}+\mathrm{x}^{2}+1}=\frac{1}{2} \int \frac{\left(\mathrm{x}^{2}+1\right)-\left(\mathrm{x}^{2}-1\right)}{\mathrm{x}^{4}+\mathrm{x}^{2}+1} \mathrm{dx}\)

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

\(=\frac{1}{2} \int \frac{1+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x-\frac{1}{2} \int \frac{1-\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x\)

= \(\frac{1}{2} \int \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+3}-\frac{1}{2} \int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)-1}\)

=> K= \(\frac{1}{2} \int \frac{\mathrm{du}}{\mathrm{u}^{2}+3}-\frac{1}{2} \int \frac{\mathrm{dv}}{\mathrm{v}^{2}-1}\)

Tìm nguyên hàm của 

a/ \(I_{5}=\int \frac{x^{3}-x+7}{2 x+5} d x\)

b/ \(\mathrm{I}_{6}=\int \frac{3 \mathrm{x}^{3}+3 \mathrm{x}^{2}+\mathrm{x}+2}{\mathrm{x}-1} \mathrm{dx}\)

c/ \(I_{7}=\int \frac{4 x^{4}+3 x^{2}+x+2}{2 x+1} d x\)

giải:

a/ Chia tử số cho mẫu số ta được

\(\frac{x^{3}-x+7}{2 x+5}=\frac{1}{2} x^{2}-\frac{5}{4} x+\frac{21}{8}-\frac{\frac{49}{8}}{2 x+5}\)

khi đó

\(I_{5}=\int \frac{x^{3}-x+7}{2 x+5} d x=\int\left(\frac{1}{2} x^{2}-\frac{5}{4} x+\frac{21}{8}-\frac{\frac{49}{8}}{2 x+5}\right) d x\)

= \(\int\left(\frac{1}{2} x^{2}-\frac{5}{4} x+\frac{21}{8}\right) d x-\frac{49}{8} \int \frac{d x}{2 x+5}\)

\(=\frac{1}{2} \cdot \frac{x^{3}}{3}-\frac{5}{4} \cdot \frac{x^{2}}{2}+\frac{21}{8} x-\frac{49}{16} \int \frac{d(2 x+5)}{2 x+5}\)

= \(\frac{x^{3}}{6}-\frac{5 x^{2}}{8}+\frac{21 x}{8}-\frac{49}{16} \ln |2 x+5|+C\)

b/ ta có \(I_{6}=\int \frac{3 x^{3}+3 x^{2}+x+2}{x-1} d x\)

= \(\int\left(3 x^{2}+6 x+7+\frac{9}{x-1}\right) d x\)

= \(x^{3}+3 x^{2}+7 x+9 \ln |x-1|+C\)

c/ chia tử số cho mẫu số ta được

\(\frac{4 x^{4}+3 x^{2}+x+2}{2 x+1}=2 x^{2}-x^{2}+2 x-\frac{1}{2}+\frac{\frac{5}{2}}{2 x+1}\)

\(I_{7}=\int \frac{4 x^{4}+3 x^{2}+x+2}{2 x+1} d x=\int\left(2 x^{3}-x^{2}+2 x-\frac{1}{2}+\frac{\frac{5}{2}}{2 x+1}\right) d x\)

= \(\int\left(2 x^{3}-x^{2}+2 x-\frac{1}{2}\right) d x+\frac{5}{2} \int \frac{d x}{2 x+1}\)

\(=2 \cdot \frac{x^{4}}{4}-\frac{x^{3}}{3}+x^{2}-\frac{1}{2} x+\frac{5}{4} \int \frac{d(2 x+1)}{2 x+1}\)

= \(\frac{x^{4}}{2}-\frac{x^{3}}{3}+x^{2}-\frac{1}{2} x+\frac{5}{4} \ln |2 x+1|+C\)