Math, asked by Anonymous, 1 month ago

Solve the integral:
\int \sqrt[x]{e}

Answers

Answered by anindyaadhikari13
4

\textsf{\large{\underline{Solution}:}}

We have to evaluate the given integral.

\displaystyle\rm=\int \sqrt[\rm x]{\rm e}\: dx

\displaystyle\rm=\int \sqrt[\rm x]{\rm e}\cdot 1\: dx

Integrating by parts, we get:

\displaystyle\rm=x\sqrt[\rm x]{\rm e}-\int-\dfrac{\sqrt[\rm x]{\rm e} }{x}\: dx

Let us assume that:

\rm:\longmapsto u=\dfrac{1}{x}

\rm:\longmapsto \dfrac{du}{dx}=\dfrac{-1}{x^{2}}

\rm:\longmapsto dx=-x^{2}\: du

Therefore, the integral becomes:

\displaystyle\rm=x\sqrt[\rm x]{\rm e}-\int\dfrac{e^{u}}{u}\: du

\displaystyle\rm=x\sqrt[\rm x]{\rm e}-Ei(u)+C

Put u = 1/x, we get:

\displaystyle\rm=x\sqrt[\rm x]{\rm e}-Ei\bigg(\dfrac{1}{x}\bigg)+C

Therefore:

\displaystyle\rm:\longmapsto\int \sqrt[\rm x]{\rm e}\: dx=x\sqrt[\rm x]{\rm e}-Ei\bigg(\dfrac{1}{x}\bigg)+C

⊕ Which is our required answer.

Note:

\displaystyle\rm:\longmapsto Ei(x)=-\int^{\infty}_{-x}\dfrac{e^{-t}}{t}\: dt=\int^{x}_{-\infty}\dfrac{e^{t}}{t}\: dt

\textsf{\large{\underline{More To Know}:}}

\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf kx+C\\ \\ \sf sin(x)&\sf-cos(x)+C\\ \\ \sf cos(x)&\sf sin(x)+C\\ \\ \sf{sec}^{2}(x)&\sf tan(x)+C\\ \\ \sf{cosec}^{2}(x)&\sf-cot(x)+C\\ \\ \sf sec(x)\  tan(x)&\sf sec(x)+C\\ \\ \sf cosec(x)\ cot(x)&\sf-cosec(x)+C\\ \\ \sf tan(x)&\sf log(sec(x))+C\\ \\ \sf\dfrac{1}{x}&\sf log(x)+C\\ \\ \sf{e}^{x}&\sf{e}^{x}+C\\ \\ \sf x^{n},n\neq-1&\sf\dfrac{x^{n+1}}{n+1}+C\end{array}}


anindyaadhikari13: Thanks for the Brainliest :)
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