Real integrals of multivalued functions via residues

$\displaystyle \int_{-i\infty }^{i\infty } \frac{dz}{(z^2-4)\log (z+1)}$ (the integration is over the imaginary axis),

$\displaystyle \int_0^\infty \frac{x^\alpha }{x^3+1}\, dx$ for $\alpha$ in the range $-1 < \alpha < 2$.

$\int_{-2}^{2} \frac{d x}{\sqrt[4]{(2+x)^{2}\left(4-x^{2}\right)}}$;

$\int_{0}^{2} \frac{d x}{\sqrt[3]{x^{2}(2-x)}}$;

$\int_{0}^{1} \frac{d x}{(x-3)^{2} \sqrt{x-x^{2}}}$;

$\int_{0}^{6} \frac{1}{x+2} \sqrt{\frac{x}{6-x}} d x$;

$\int_{0}^{5} \frac{\sqrt[4]{x(5-x)^{7}}}{x^{2}-5 x-6} d x$;

$\int_{-1}^{1} \frac{x \sqrt{1-x^{2}}}{x^{2}-3 x+2} d x$;

$\int_{0}^{3} \frac{d x}{(x+2)^{2} \sqrt[4]{x^{3}(3-x)}}$;

$\int_{0}^{2} \frac{\sqrt[3]{x^{2}(2-x)}}{(x+2)^{2}} d x$;

$\int_{0}^{3} \sqrt[3]{(3-x) x^{2}} d x$;

$\int_{0}^{1} \sqrt{x-x^{2}} \frac{20 x+39}{x+2} d x$;

$\int_{-1}^{1} \sqrt[4]{\frac{1+x}{1-x}} \frac{3 x^{2}+13 x+20}{(x+2)^{2}} d x$;

$\int_{-1}^{0} \sqrt{-1-\frac{1}{x}} \frac{3 x+1}{3 x+4} d x$.

$\int_{1}^{+\infty } \frac{\sqrt{x-1} \ln (x-1)}{x^{2}+3 x} d x$;

$\int_{2}^{+\infty } \frac{\ln (x-2)}{\left(x^{2}-1\right) \sqrt{x-2}} d x$;

$\int_{0}^{+\infty } \frac{\sqrt{x} \ln x d x}{x^{2}+1}$;

$\int_{0}^{+\infty } \frac{\ln x}{(x+1) \sqrt[4]{x^{3}}} d x$;

$\int_{0}^{+\infty } \frac{\ln x d x}{(x+1)(2 x+1) \sqrt{x}}$;

$\int_{1}^{+\infty } \frac{d x}{\left(x^{2}+1\right) \sqrt{x-1}}$;

$\int_{0}^{+\infty } \frac{d x}{\sqrt[5]{x}(x+1)^{2}}$;

$\int_{0}^{+\infty } \frac{\sqrt{x} \ln x}{(x+1)(x+2)} d x$.

$\int_{0}^{+\infty } \frac{\ln x}{(x+1)^{2}} d x$;

$\int_{0}^{+\infty } \frac{\ln x d x}{x^{2}+2 x+2}$;

$\int_{0}^{+\infty } \frac{\ln x d x}{\left(x^{2}+1\right)^{2}}$;

$\int_{0}^{+\infty } \frac{\ln x d x}{(x+1)\left(x^{2}+1\right)}$;

$\int_{0}^{+\infty } \frac{x \ln x}{\left(x^{2}+1\right)^{2}} d x$;

$\int_{0}^{+\infty } \frac{x^{2} \ln x}{\left(x^{2}+1\right)^{2}} d x$;

$\int_{0}^{+\infty } \frac{\ln x}{x^{2}-1} d x$;

$\int_{0}^{+\infty }\left(\frac{\ln x}{x-1}\right)^{2} d x$;

$\int_{0}^{+\infty } \frac{\ln^{2} x}{\left(x^{2}+1\right)^{2}} d x$;

$\int_{0}^{+\infty }\left(\frac{\ln x}{x-1}\right)^{3} d x$;

$\int_{-\infty }^{+\infty } \frac{x}{\operatorname {sh} x} d x$;

$\int_{0}^{+\infty } \frac{x d x}{\operatorname {sh} x \operatorname {ch} x}$;

$\int_{0}^{+\infty }\left(\frac{x}{\operatorname {sh} x}\right)^{2} d x$;

$\int_{0}^{+\infty }\left(\frac{x}{\operatorname {sh} x}\right)^{3} d x$;

$\int_{0}^{+\infty } \frac{x^{2}}{\operatorname {ch} x} d x$;

$\int_{0}^{1} \ln \frac{1-x}{x} \frac{d x}{1+x^{2}}$;

$\int_{0}^{1}\left(\ln \frac{x}{1-x}\right)^{2} \frac{d x}{x+1}$;

$\int_{-1}^{1}\left(\ln \frac{1+x}{1-x}\right)^{2} \frac{d x}{x^{2}}$;

$\int_{0}^{+\infty } \frac{d x}{\left(x^{2}+\pi^{2}\right) \operatorname {ch} x}$;

$\int_{-\infty }^{+\infty } \frac{e^{x} d x}{\left(x^{2}+\pi^{2}\right) \operatorname {ch} x}$;

$\int_{0}^{+\infty } \frac{d x}{\left(x^{2}+9 \pi^{2}\right) \operatorname {ch} x}$;

$\int_{0}^{+\infty } \frac{d x}{(x+1)\left(\ln^{2} x+\pi^{2}\right)}$;

$\int_{-1}^{1} \frac{d x}{\left(\ln \frac{1+x}{1-x}\right)^{2}+9 \pi^{2}}$;

$\int_{0}^{1} \frac{x d x}{\left(\ln \frac{x}{1-x}\right)^{2}+\pi^{2}}$;

$\int_{0}^{+\infty } \frac{\ln x d x}{x^{2}+2 a x \cos \lambda +a^{2}}, a > 0, a < \lambda < \pi$.

$\int_{-\infty }^{+\infty } R(x) \ln \left|x-a\right| d x = -2 \pi \mathrm{Im}\left(\sum_{s = 1}^{n}\right) \mathop{\mathrm{res}}_{z = a_{s}} R(z) g(z)$ (here $a$ is a real number, and $g(z)$ is an arbitrary branch of the multivalued function $\operatorname {Ln}(z-a)$, regular in the upper half-plane);

$\int_{-\infty }^{+\infty } R(x)\left|x-a\right|^{\alpha -1} d x = -\frac{2 \pi }{\sin \frac{\pi \alpha }{2}} \mathrm{Im}\left(\sum_{s = 1}^{n}\right) \mathop{\mathrm{res}}_{z = a_{s}} R(z) g(z)$ (here $a$ is a real number, $\alpha > 0$, and $g(z)$ is the branch of the multivalued function $\left(\frac{z-a}{i}\right)^{\alpha -1}$, regular in the upper half-plane and positive for positive values of $\frac{z-a}{i}$);

$\int_{-\infty }^{+\infty } R(x) \ln \left|x^{2}-a^{2}\right| d x = -2 \pi \mathrm{Im}\left(\sum_{s = 1}^{n}\right) \mathop{\mathrm{res}}_{z = a_{s}} R(z) g(z)$ (here $a > 0$, and $g(z)$ is an arbitrary branch of the multivalued function $\operatorname {Ln}\left(z^{2}-a^{2}\right)$, regular in the upper half-plane);

$\int_{-\infty }^{+\infty } R(x) \ln \left(x^{2}+a^{2}\right) d x = -4 \pi \mathrm{Im}\left(\sum_{s = 1}^{n}\right) \mathop{\mathrm{res}}_{z = a_{s}} R(z) g(z)$ (here $a > 0$, and $g(z)$ is an arbitrary branch of the multivalued function $\operatorname {Ln}(z+a i)$, regular in the upper half-plane).

$\int_{-\infty }^{+\infty } \frac{\sqrt[3]{\left|x-1\right|}}{x^{2}+1} d x$;

$\int_{-\infty }^{+\infty } \frac{\ln \left|x^{2}-1\right|}{x^{2}+1} d x$;

$\int_{-\infty }^{+\infty } \frac{\ln \left(x^{4}+2 x^{2}+2\right)}{x^{2}+4} d x$;

$\int_{-\infty }^{+\infty } \sqrt{\sqrt{x^{2}+1}-x} \frac{d x}{x^{2}+1}$;

$\int_{-\infty }^{+\infty } \frac{d x}{\left(x^{2}+1\right) \sqrt{\left|x-1\right|}}$;

$\int_{0}^{+\infty } \frac{\ln \left(x^{2}+2 x+2\right)}{\left(x^{2}+1\right)^{2}} d x$;

$\int_{0}^{+\infty } \frac{\ln \left(1+x^{2}\right)}{x^{2}} \frac{d x}{1+x^{2}}$;

$\int_{0}^{+\infty } \frac{\operatorname {arctg} x}{x} \frac{d x}{1+x^{2}}$;

$\int_{0}^{+\infty }(\operatorname {arctg} x)^{2} \frac{d x}{x^{2}}$;

$\int_{-\infty }^{+\infty } \frac{\operatorname {arctg} x}{x^{2}-4 x+5} \ln \left(x^{2}+1\right) d x$;

$\int_{0}^{+\infty } \frac{\sqrt{\sqrt{x^{2}+1}+x}-\sqrt{\sqrt{x^{2}+1}-x}}{x\left(x^{2}+1\right)} d x$;

$\int_{0}^{+\infty }\left(\sqrt{\sqrt{1+x^{2}}+x}+\sqrt{\sqrt{1+x^{2}}-x}-2\right) \frac{d x}{x^{2}}$;

$\int_{0}^{+\infty } \frac{\sqrt{\sqrt{1+x^{2}}+x}-\sqrt{\sqrt{1+x^{2}}-x}}{x \sqrt{1+x^{2}}} d x$;

$\int_{0}^{+\infty } \sqrt{\sqrt{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}+x^{2}-a b} \frac{d x}{\left(x^{2}+1\right)^{2}}, a > 0, b > 0$.