indefinite-integration

Related Questions

• $\int_{{}}^{{}}{\frac{dx}{{{e}^{-2x}}{{({{e}^{2x}}+1)}^{2}}}=}$

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• To find the value of $\int_{{}}^{{}}{\frac{1+\log x}{x}\text{ }}dx$, the proper substitution is

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• $\int_{{}}^{{}}{2x{{\cos }^{3}}{{x}^{2}}\sin {{x}^{2}}dx=}$

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• $\int_{{}}^{{}}{\frac{x}{\sqrt{4-{{x}^{4}}}}dx}=$

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• $\int_{{}}^{{}}{\frac{{{e}^{2x}}+1}{{{e}^{2x}}-1}\ dx}$ equals

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