\(\displaystyle{9}{\left({9}\right)}+{1.5}{\left({9}^{{{2}}}\right)}−{9}{\left({7}\right)}−{1.5}{\left({7}^{{{2}}}\right)}={66}\)

Here you go.

Here you go.

asked 2021-09-11

\(\int_{9}^{4} \bigg(\sqrt x+\frac{1}{\sqrt x}\bigg)^{2}dx\)

asked 2022-01-05

How to compute

\(\displaystyle{\int_{{-\infty}}^{\infty}}{\exp{{\left(-{\frac{{{\left({x}^{{2}}-{13}{x}-{1}\right)}^{{2}}}}{{{611}{x}^{{2}}}}}\right)}}}{\left.{d}{x}\right.}\)

\(\displaystyle{\int_{{-\infty}}^{\infty}}{\exp{{\left(-{\frac{{{\left({x}^{{2}}-{13}{x}-{1}\right)}^{{2}}}}{{{611}{x}^{{2}}}}}\right)}}}{\left.{d}{x}\right.}\)

asked 2022-01-04

I want to evaluate the following integral via complex analysis

\(\displaystyle{\int_{{{x}={0}}}^{{{x}=\infty}}}{e}^{{-{a}{x}}}{\cos{{\left({b}{x}\right)}}}{\left.{d}{x}\right.},\ {a}{>}{0}\)

Which function/ contour should I consider ?

\(\displaystyle{\int_{{{x}={0}}}^{{{x}=\infty}}}{e}^{{-{a}{x}}}{\cos{{\left({b}{x}\right)}}}{\left.{d}{x}\right.},\ {a}{>}{0}\)

Which function/ contour should I consider ?

asked 2022-01-05

Evaluate the integral \(\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{\cos{{\left({x}\right)}}}}}{{{x}^{{2}}+{1}}}}{\left.{d}{x}\right.}\)

asked 2022-01-03

How can I evaluate

\(\displaystyle{\int_{{-\infty}}^{\infty}}{\frac{{{\cos{{x}}}}}{{{\text{cosh}{{x}}}}}}{\left.{d}{x}\right.}\) and \(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{\sin{{x}}}}}{{{e}^{{x}}-{1}}}}{\left.{d}{x}\right.}\)

\(\displaystyle{\int_{{-\infty}}^{\infty}}{\frac{{{\cos{{x}}}}}{{{\text{cosh}{{x}}}}}}{\left.{d}{x}\right.}\) and \(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{\sin{{x}}}}}{{{e}^{{x}}-{1}}}}{\left.{d}{x}\right.}\)

asked 2022-01-03

\(\displaystyle{\int_{{-\infty}}^{\infty}}{e}^{{-{\frac{{{1}}}{{{2}}}}{x}^{{2}}}}{\left.{d}{x}\right.}\) and \(\displaystyle{\int_{{-\infty}}^{\infty}}{x}^{{2}}{e}^{{-{\frac{{{1}}}{{{2}}}}{x}^{{2}}}}{\left.{d}{x}\right.}\)

how i compute these integrals via Gauss Integral?

how i compute these integrals via Gauss Integral?

asked 2021-12-30

How to evaluate:

\(\displaystyle{\int_{{-\infty}}^{\infty}}{\frac{{{\ln{{\left({x}^{{2}}+{1}\right)}}}}}{{{x}^{{2}}+{1}}}}{\left.{d}{x}\right.}\)

\(\displaystyle{\int_{{-\infty}}^{\infty}}{\frac{{{\ln{{\left({x}^{{2}}+{1}\right)}}}}}{{{x}^{{2}}+{1}}}}{\left.{d}{x}\right.}\)