Math, asked by Anonymous, 1 month ago

Solve the integral,
 \int x +1 dx

Answers

Answered by anindyaadhikari13
4

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

We have to integrate the given function.

 \displaystyle =\rm \int(x + 1) \: dx

We know that:

 \displaystyle\rm: \longmapsto\int f(x) \pm g(x) \: dx =  \int f(x) \: dx + \int g(x) \: dx

Therefore, we get:

 \displaystyle =\rm \int x\: dx +  \int 1 \: dx

We know that:

 \displaystyle =\rm \int  {x}^{n} \: dx=\dfrac{x^{n+1}}{n+1}+C

Therefore, we get:

 \displaystyle =\rm  \dfrac{ {x}^{1 + 1} }{1 + 1}  +  \dfrac{ {x}^{0 + 1} }{0 + 1}  + C

 \displaystyle =\rm  \dfrac{ {x}^{2} }{2}  +  x  + C

Therefore:

 \displaystyle\rm: \longmapsto\int (x + 1)\:  dx =  \dfrac{ {x}^{2} }{2}  + x + C

★ Which is our required answer.

\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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