Example. Evaluate the definite integral

$$
\int_{0}^{1} \frac{x d x}{x^{4}+1}
$$
Solution.

1. Let's represent the integrand as a product of two functions $F(x) g(x)$, where $g(x)$ has an obvious antiderivative $G(x)$, and $F(x)$ is a function of this antiderivative, i.e., $F(x)=u(G(x))$.

In this case,

$$
F(x)=\frac{1}{2} \frac{1}{1+\left(x^{2}\right)^{2}}, \quad g(x)=2 x, \quad G(x)=x^{2}, \quad F(x)=\frac{1}{2} \frac{1}{1+G^{2}}=u(G)
$$

2. Then

$$
\int_{0}^{1} \frac{x d x}{x^{4}+1}=\frac{1}{2} \int_{0}^{1} \frac{1}{1+\left(x^{2}\right)^{2}} d x^{2}=\frac{1}{2} \int_{0}^{1} \frac{1}{1+G^{2}} d G
$$

where $G=x^{2}$ and $G(0)=0, G(1)=1$.

3. The last integral is a standard one. Applying the Newton-Leibniz formula:

$$
\frac{1}{2} \int_{0}^{1} \frac{1}{1+G^{2}} d G=\left.\frac{1}{2} \operatorname{arctg} G\right|_{0} ^{1}=\frac{\pi}{8}
$$

Answer. $\int_{0}^{1} \frac{x d x}{x^{4}+1}=\frac{\pi}{8}$.

Conditions of the Problems. Calculate the definite integrals.

1. $\int_{0}^{1} \frac{x}{\left(x^{2}+1\right)^{2}} d x$.
2. $\int_{0}^{\pi / 4} \frac{\sin x-\cos x}{(\cos x+\sin x)^{3}} d x$.
3. $\int_{0}^{1} \frac{x^{3}+1}{\left(x^{4}+4 x+2\right)^{2}} d x$.
4. $\int_{1}^{e} \frac{1+\ln ^{3} x}{x} d x$.
5. $\int_{\pi / 4}^{\pi / 2} \frac{x \cos x+\sin x}{(x \sin x)^{3}} d x$.
6. $\int_{1}^{4} \frac{3 \sqrt{x}+1}{2 x \sqrt{x}+x} d x$.
7. $\int_{0}^{\pi / 4} \frac{2 \operatorname{tg} x}{\cos ^{2} x} d x$.

$\sqrt{2} / 2$

6. $\int_{0} \frac{x}{\sqrt{1-x^{4}}} d x$.
7. $\int_{0}^{1} \frac{2 \operatorname{arctg} x+x}{1+x^{2}} d x$.

$1 / 2$

10. $\int_{0} \frac{2 \arcsin x+x}{\sqrt{1-x^{2}}} d x$.

Answers. 1.1/4. 2. $-1 / 4$. 3.5/56. 4.1. 5.5/4. 6. $\pi / 12$. $7.14 / \pi^{2}$. 8. $\pi^{2} / 16+\ln \sqrt{2} . \quad 9 . \ln (20 / 3) . \quad 10 .\left(\pi^{2}-18 \sqrt{3}+36\right) / 36$.

## 8.2. Integration by Parts

PROBLEM STATEMENT. Calculate the definite integral

$$
\int_{a}^{b} F(x) g(x) d x
$$

PLAN OF SOLUTION. Suppose on the interval $[a, b]$ the function $g(x)$ has an obvious antiderivative $G(x)$, and $F(x)$ is a differentiable function, with its derivative $f(x)=F^{\prime}(x)$ being a simpler function than $F(x)$. Then we apply the integration by parts formula

$$
\int_{a}^{b} F(x) g(x) d x=\left.F(x) G(x)\right|_{a} ^{b}-\int_{a}^{b} f(x) G(x) d x
$$

If the method is chosen correctly, the integral on the right side of this equation turns out to be a standard one or can be reduced to a standard one, for example, by repeated integration by parts.