\[\int {x^n dx} = {x^{n+1} \over n+1}, \quad n \neq -1\]

\[\int \sqrt{x}\,dx = \frac23 x^{3/2}\]

\[\int {\sqrt{x}\,dx \over x+a} = 2\sqrt{x} - 2\sqrt{a}\,\mathtt{arctg}\sqrt{\frac{x}{a}}, \quad a \gt 0\]

\[\int {\sqrt{x}\,dx \over x-a} = 2\sqrt{x} - \sqrt{a}\,\ln\left|{\sqrt{x}+\sqrt{a} \over \sqrt{x}-\sqrt{a}}\right|, \quad a \gt 0\]

\[\int {\sqrt{x}\,dx \over (x+a)^2} = - {\sqrt{x} \over x+a} + \frac{1}{\sqrt{a}} \mathtt{arctg}\sqrt{\frac{x}{a}}, \quad a \gt 0\]

\[\int {\sqrt{x}\,dx \over (x-a)^2} = - {\sqrt{x} \over x-a} - \frac{1}{2\sqrt{2}} \ln\left|{\sqrt{x}+\sqrt{a} \over \sqrt{x}-\sqrt{a}}\right|, \quad a \gt 0\]

\[\int {dx \over \sqrt{x}(x+a)} = \frac{2}{\sqrt{a}} \mathtt{arctg}\sqrt{\frac{x}{a}}, \quad a \gt 0\]

\[\int {dx \over \sqrt{x}(x-a)} = - \frac{1}{\sqrt{a}} \ln\left|{\sqrt{x}+\sqrt{a} \over \sqrt{x}-\sqrt{a}}\right|, \quad a \gt 0\]

\[\int {x\sqrt{x}\,dx \over x+a} = - 2a\sqrt{x} + \frac23 x\sqrt{x} + 2a\sqrt{a}\,\mathtt{arctg}\sqrt{\frac{x}{a}}, \quad a \gt 0\]

\[\int {dx \over \sqrt{x}(x-a)^2} = - {\sqrt{x} \over a(x-a)} + \frac{1}{2a\sqrt{a}} \ln\left|{\sqrt{x}+\sqrt{a} \over \sqrt{x}-\sqrt{a}}\right|, \quad a \gt 0\]

\[\int {dx \over \sqrt{x}(x+a)^2} = {\sqrt{x} \over a(x+a)} + \frac{1}{a\sqrt{a}} \mathtt{arctg}\sqrt{\frac{x}{a}}, \quad a \gt 0\]

\[\int {dx \over x\sqrt{x}(x+a)} = - {2 \over a\sqrt{x}} - {2 \over a\sqrt{a}} \mathtt{arctg}\sqrt{\frac{x}{a}}, \quad a \gt 0\]

\[\int {dx \over x\sqrt{x}(x-a)} = {2 \over a\sqrt{x}} - {1 \over a\sqrt{a}} \ln\left|{\sqrt{x}+\sqrt{a} \over \sqrt{x}-\sqrt{a}}\right|, \quad a \gt 0\]

\[\int {\sqrt{x} dx \over x^2+a^2} = - {1 \over 2\sqrt{2a}} \ln{x+\sqrt{2xa}+a \over x-\sqrt{2xa}+a } - \frac{1}{\sqrt{2a}} \mathtt{arctg}\frac{\sqrt{2ax}}{x-a}, \quad a \gt 0\]

\[\int {\sqrt{x} dx \over x^2-a^2} = - {1 \over 2\sqrt{a}} \ln\left|{\sqrt{x}+\sqrt{a} \over \sqrt{x}-\sqrt{a}}\right| + \frac{1}{\sqrt{a}} \mathtt{arctg}\sqrt{\frac{x}{a}}, \quad a \gt 0\]

\[\int {dx \over \sqrt{x}(x^2+a^2)} = {1 \over 2a\sqrt{2a}} \ln\left|{x+\sqrt{2xa}+a \over x-\sqrt{2xa}+a }\right| + \frac{1}{a\sqrt{2a}} \mathtt{arctg}\frac{\sqrt{2x}+\sqrt{a}}{\sqrt{a}} + \frac{1}{a\sqrt{2a}} \mathtt{arctg}\frac{\sqrt{2x}-\sqrt{a}}{\sqrt{a}}, \quad a \gt 0\]

\[\int {dx \over \sqrt{x}(x^2-a^2)} = - {1 \over 2a\sqrt{a}} \ln\left|{\sqrt{x}+\sqrt{a} \over \sqrt{x}-\sqrt{a}}\right| - {1 \over a\sqrt{a}} \mathtt{arctg}\sqrt{\frac{x}{a}}, \quad a \gt 0\]

\[\int {dx \over \sqrt{x+a}} = 2\sqrt{x+a}\]

\[\int {dx \over (x+a)^{3/2}} = - {2 \over \sqrt{x+a}}\]

\[\int {xdx \over \sqrt{x+a}} = \frac23 (x+a)^{3/2} - 2a\sqrt{x+a}\]

\[\int {xdx \over (x+a)^{3/2}} = 2\sqrt{x+a} + {2a \over \sqrt{x+a}}\]

\[\int {dx \over x\sqrt{x+a}} = \frac{1}{\sqrt{a}} \ln\left|{\sqrt{x+a}-\sqrt{a} \over \sqrt{x+a}+\sqrt{a}}\right|, \quad a \gt 0\]

\[\int {dx \over x\sqrt{x-a}} = \frac{2}{\sqrt{a}} \mathtt{arctg}\sqrt{\frac{x-a}{a}}, \quad a \gt 0\]

\[\int {dx \over x\sqrt{a-x}} = - \frac{1}{\sqrt{a}} \ln\left|{\sqrt{a-x}+\sqrt{a} \over \sqrt{a-x}-\sqrt{a}}\right|, \quad a \gt 0\]

\[\int \sqrt{x+a}\,dx = \frac23 (x+a)^{3/2}\]

\[\int x\sqrt{x+a}\,dx = \frac25 (x+a)^{5/2} - \frac{2a}{3} (x+a)^{3/2}\]

\[\int {\sqrt{x+a}\over \sqrt{x+b}}dx = (x+b){\sqrt{x+a}\over \sqrt{x+b}} + \frac{a-b}{2} \ln\left|{1+\sqrt{(x+a)/(x+b)} \over 1-\sqrt{(x+a)/(x+b)} }\right|\]

\[\int {\sqrt{x+a}\over \sqrt{b-x}}dx = (x-b){\sqrt{x+a}\over \sqrt{b-x}} + (a+b) \mathtt{arctg}\sqrt{{x+a \over b-x}}\]

\[\int {dx \over \sqrt{x^2+a^2}} = \ln(x + \sqrt{x^2+a^2}), \quad a \neq 0\]

\[\int {dx \over (x^2+a^2)^{3/2}} = {1 \over a^2}{x \over \sqrt{x^2+a^2}}, \quad a \neq 0\]

\[\int {xdx \over \sqrt{x^2+a^2}} = \sqrt{x^2+a^2}\]

\[\int {xdx \over (x^2+a^2)^{3/2}} = - {1 \over \sqrt{x^2+a^2}}\]

\[\int {xdx \over (x^2+a^2)^{5/2}} = - {1 \over 3(x^2+a^2)^{3/2}}\]

\[\int {x^2 dx \over \sqrt{x^2+a^2}} = \frac{x}{2} \sqrt{x^2+a^2} - \frac{a^2}{2} \ln(x + \sqrt{x^2+a^2})\]

\[\int {x^2 dx \over (x^2+a^2)^{3/2}} = - {x \over \sqrt{x^2+a^2}} + \ln(x + \sqrt{x^2+a^2})\]

\[\int {x^3 dx \over \sqrt{x^2+a^2}} = \frac13 (x^2+a^2)^{3/2} - a^2\sqrt{x^2+a^2}\]

\[\int {dx \over x\sqrt{x^2+a^2}} = - \frac{1}{a} \ln\left|{a+\sqrt{x^2+a^2} \over x}\right|, \quad a \neq 0\]

\[\int {dx \over x^2\sqrt{x^2+a^2}} = - {\sqrt{x^2+a^2} \over a^2x}, \quad a \neq 0\]

\[\int {dx \over x^3\sqrt{x^2+a^2}} = - {\sqrt{x^2+a^2} \over 2a^2x^2} + \frac{1}{2a^3} \ln\left|{a+\sqrt{x^2+a^2} \over x}\right|, \quad a \neq 0\]

\[\int \sqrt{x^2+a^2}\,dx = \frac{x}{2} \sqrt{x^2+a^2} + \frac{a^2}{2} \ln|x+\sqrt{x^2+a^2}|\]

\[\int (x^2+a^2)^{3/2}\,dx = \frac18 \sqrt{x^2+a^2} (2x^3+5a^2x) + \frac38 a^4 \ln(x+\sqrt{x^2+a^2})\]

\[\int x\sqrt{x^2+a^2}\,dx = \frac13 (x^2+a^2)^{3/2}\]

\[\int x(x^2+a^2)^{3/2}\,dx = \frac15 (x^2+a^2)^{5/2}\]

\[\int x^2\sqrt{x^2+a^2}\,dx = {2x^3+a^2x \over 8}\sqrt{x^2+a^2} - \frac{a^4}{8} \ln(x+\sqrt{x^2+a^2})\]

\[\int {\sqrt{x^2+a^2} \over x} dx = \sqrt{x^2+a^2} - a\ln\left|{a+\sqrt{x^2+a^2} \over x}\right|\]

\[\int {(x^2+a^2)^{3/2} \over x} dx = \frac13 (x^2+a^2)^{3/2} + a^2\sqrt{x^2+a^2} - a^3\ln\left|{a+\sqrt{x^2+a^2} \over x}\right|\]

\[\int {\sqrt{x^2+a^2} \over x^2} dx = - {\sqrt{x^2+a^2} \over x} + \ln(x+\sqrt{x^2+a^2})\]

\[\int {\sqrt{x^2+a^2} \over x^3} dx = - {\sqrt{x^2+a^2} \over 2x^2} - \frac{1}{2a} \ln\left|{a+\sqrt{x^2+a^2} \over x}\right|, \quad a \neq 0\]

\[\int {dx \over \sqrt{x^2 \pm a^2}} = \ln|x+\sqrt{x^2 \pm a^2}|\]

\[\int {dx \over (x^2-a^2)^{3/2}} = - {1 \over a^2}{x \over \sqrt{x^2-a^2}}, \quad a \neq 0\]

\[\int {xdx \over \sqrt{x^2-a^2}} = \sqrt{x^2-a^2}\]

\[\int {xdx \over (x^2-a^2)^{3/2}} = - {1 \over \sqrt{x^2-a^2}}\]

\[\int {xdx \over (x^2-a^2)^{5/2}} = - {1 \over 3(x^2-a^2)^{3/2}}\]

\[\int {x^2 dx \over \sqrt{x^2-a^2}} = {x\sqrt{x^2-a^2} \over 2} + \frac{a^2}{2} \ln|x+\sqrt{x^2-a^2}|\]

\[\int {x^2 dx \over (x^2-a^2)^{3/2}} = - {x \over \sqrt{x^2-a^2}} + \ln|x+\sqrt{x^2-a^2}|\]

\[\int {x^3 dx \over \sqrt{x^2-a^2}} = \frac13 (x^2-a^2)^{3/2} + a^2\sqrt{x^2-a^2}\]

\[\int {dx \over x\sqrt{x^2-a^2}} = - \frac{1}{a} \mathtt{arcsin}|\frac{a}{x}|, \quad a \gt 0\]

\[\int {dx \over x^2\sqrt{x^2-a^2}} = {\sqrt{x^2-a^2} \over a^2x}, \quad a \neq 0\]

\[\int {dx \over x^3\sqrt{x^2-a^2}} = {\sqrt{x^2-a^2} \over 2a^2x^2} \mathtt{arcsin}|\frac{a}{x}|, \quad a \gt 0\]