Example 4. Find $\int \frac{d x}{x^{4}-3 x^{3}+2 x^{2}}$.
Solution. The denominator can be factored as: $x^{2}\left(x^{2}-3 x+2\right)=x^{2}(x-1)(x-2)$. The integrand fraction is equal to the sum of four type I fractions.

$$
\begin{aligned}
& \frac{1}{x^{2}(x-1)(x-2)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}+\frac{D}{x-2}= \\
& \quad=\frac{A x(x-1)(x-2)+B(x-1)(x-2)+C x^{2}(x-2)+D x^{2}(x-1)}{x(x-1)(x-2)}
\end{aligned}
$$

The coefficients $A, B, C, D$ are found by the following substitution principle:

1) in the numerators of the right and left parts of the obtained equality, we substitute $x=0 \cdot 1=B \cdot(-1)(-2)$, hence $B=\frac{1}{2}$,
2) in the same numerators, we substitute $x=1: 1=C \cdot(-1)$, hence $C=-1$,
3) now we substitute $x=2 \cdot 1=D \cdot 4$, we find $D=\frac{1}{4}$

There are no more such convenient substitutions, and the coefficient $A$ remains undefined. We will find it the old way, for example, by equating the coefficients of the highest degree $x^{3}$ in the numerators of the right and left parts of the equality $0=A+C+D$. Hence $A=-C-D=1-\frac{1}{4}=\frac{3}{4}$. Therefore,

$\int \frac{d x}{x^{4}-3 x^{3}+2 x^{2}}=\frac{3}{4} \int \frac{d x}{x}+\frac{1}{2} \int \frac{d x}{x^{2}}-\int \frac{d x}{x-1}+\frac{1}{4} \int \frac{d x}{x-2}=$ $=\frac{3}{4} \ln |x|-\frac{1}{2 x}-\ln |x-1|+\ln |x-2|+C=-\frac{1}{2 x}+\frac{1}{4} \ln \frac{\left|x^{3}(x-2)\right|}{(x-1)^{4}}+C$.

## Exercises

Find the integrals.

1. $\int \frac{x^{3}+1}{x(x+1)^{3}} d x$.
2. $\int \frac{d x}{(x+1)(x+2)^{2}(x+3)^{2}}$.
3. $\int \frac{6 x-4}{x^{3}-4 x} d x$.
4. $\int \frac{d x}{x^{2}-6 x+5}$.
5. $\int \frac{d x}{x^{3}+x^{2}}$.
6. $\int \frac{2 x^{2}-3 x+3}{x^{3}-2 x^{2}+x} d x$.
7. $\int \frac{x^{3}+3 x^{2}+5 x+7}{x^{2}+2} d x$
8. $\int \frac{d x}{(x+1)\left(x^{2}+1\right)}$.
9. $\int \frac{2 x^{3}+3 x}{x^{4}+x^{2}+1} d x$.
10. $\int \frac{3 x+2}{\left(x^{2}+2 x+10\right)^{2}} d x$.
11. $\int \frac{x^{2} d x}{x^{6}+2 x^{3}+3}$.
12. $\int \frac{x^{2}+2 x+6}{(x-1)(x-2)(x-4)} d x$.
13. $\int \frac{x^{2}+1}{(x-1)^{3}(x+3)} d x$.
14. $\int \frac{d x}{x^{5}-x^{2}}$.
15. $\int \frac{x^{5}}{\left(x^{3}+1\right)\left(x^{3}+8\right)} d x$.

## Answers

1. $-\ln |x|+2 \ln |x-1|-\frac{1}{x-1}-\frac{1}{(x-1)^{2}}+C$.
2. $\frac{1}{4} \ln |x+1|+\ln |x+2|-\frac{5}{4} \ln |x+3|+\frac{1}{x+2}+\frac{1}{2(x+3)}+C$.
3. $\ln |x|+\ln |x-2|-2 \ln |x+2|+C .4 . \frac{1}{4} \ln \left|\frac{x-5}{x-1}\right|+C$.
4. $\ln \left|\frac{x+1}{x}\right|+\frac{1}{x}+C .6 .3 \ln |x|-\ln |x-1|-\frac{2}{x-1}+C$.
5. $\frac{x^{2}}{2}+3 x+\frac{3}{2} \ln \left(x^{2}+1\right)+\frac{1}{\sqrt{2}} \operatorname{arctg} \frac{x}{\sqrt{2}}+C$.
6. $\frac{1}{2} \ln |x+1|-\frac{1}{4} \ln \left(x^{2}+1\right)+\frac{1}{2} \operatorname{arctg} x+C$.
7. $\frac{1}{2} \ln \left(x^{4}+x^{2}+1\right)+\frac{2}{\sqrt{3}} \operatorname{arctg} \frac{2 x^{2}+1}{\sqrt{3}}+C$.
8. $\frac{1}{2} \ln |x+1|-\frac{1}{4} \ln \left(x^{2}+1\right)+\frac{1}{2} \operatorname{arctg} x+C$.
9. $\frac{1}{2} \ln \left(x^{4}+x^{2}+1\right)+\frac{2}{\sqrt{3}} \operatorname{arctg} \frac{2 x^{2}+1}{\sqrt{3}}+C$.
10. $-\frac{3}{2\left(x^{2}+2 x+10\right)}-\frac{1}{18} \cdot \frac{x+1}{x^{2}+2 x+10}-\frac{1}{54} \operatorname{arctg} \frac{x+1}{3}+C$.
11. $\frac{1}{3 \sqrt{2}} \operatorname{arctg} \frac{x^{3}+1}{\sqrt{2}}+C$. 12. $\ln \left|\frac{(x-1)^{3}(x-4)^{5}}{(x-2)^{7}}\right|+C$.
12. $-\frac{1}{4(x-1)^{2}}-\frac{3}{8(x-1)}+\frac{5}{32} \ln \left|\frac{x-1}{x+3}\right|+C$. 14. $\frac{1}{x}+\frac{1}{6} \ln \frac{(x-1)^{2}}{x^{2}+x+1}+$ $+\frac{1}{\sqrt{3}} \operatorname{arctg} \frac{2 x+1}{\sqrt{3}}+C .15 \cdot \frac{1}{21}\left(8 \ln \left|x^{3}+1\right|-\ln \left|x^{3}+1\right|\right)+C$.

## § 5. Integration of Trigonometric Functions

$1^{\circ} . \quad \mathrm{K}$ integrals of the form $\int \sin a x \sin b x d x, \quad \int \sin a x \cos b x d x$, $\int \cos a x \cos b x d x$ should apply the product-to-sum formulas:

$$
\begin{aligned}
\sin \alpha \cdot \sin \beta & =\frac{1}{2}[\cos (\alpha-\beta)-\cos (\alpha+\beta)] \\
\cos \alpha \cdot \cos \beta & =\frac{1}{2}[\cos (\alpha-\beta)+\cos (\alpha+\beta)] \\
\sin \alpha \cdot \cos \beta & =\frac{1}{2}[\sin (\alpha-\beta)+\sin (\alpha+\beta)]
\end{aligned}
$$

$2^{\circ}$. For integrals of the form $\int \sin ^{2 k} a x d x, \int \cos ^{2 k} a x d x$, the formulas for reducing the power should be applied:

$$
\sin ^{2} \alpha=\frac{1-\cos 2 \alpha}{2}, \quad \cos ^{2} \alpha=\frac{1+\cos 2 \alpha}{2}
$$

$3^{\circ}$. Integrals of the form $\int \sin ^{2 k+1} a x d x, \int \cos ^{2 k+1} a x d x$ are taken directly by bringing under the differential sign:

$$
\begin{aligned}
& \int \sin ^{2 k+