Question

Prove that For all n belongs to The set of Real numbers except " - 1 " :-

$\quad \qquad { \bigstar { \underline { \boxed { \displaystyle { \sf { \int {x}^{n} dx = \dfrac{{x}^{n + 1 }}{n + 1 } + C } } } } } { \bigstar } }$ ​

Answer 1

$\large \dag$ Question :-

Prove that For all n belongs to The set of Real numbers except " - 1 " :-

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

$\large \dag$ Proof :-

We Know that,

$\rm \frac{d}{dx} \bigg( \frac{x {}^{n + 1} }{n + 1} \bigg)$

$= \rm \bigg(\frac{1}{n + 1} \bigg) .\frac{d}{dx} \bigg( {x}^{n + 1} \bigg)$

$= \rm \bigg(\frac{1}{ \cancel{n + 1}} \bigg).\big( \cancel{n + 1} \big). {x}^{n}$

$= \large \bf \green{{x}^{n} }$

Therefore, definition of anti derivative or integral we can say that :

$\blue \bigstar \: \: \red{\bf\int {x}^{n} dx = \dfrac{{x}^{n + 1 }}{n + 1 } + C,x\ne1}$

$\red \dashrightarrow\large\underline{\pink{\underline{\frak{\pmb{\text Hence\:\:Proved }}}}} {}^{ \orange\bigstar \purple\bigstar}$

$\large \dag$ Additional Information :-

$\blue{ \bigstar \:   \underline{ \pmb{ \mathfrak{ Commonl \text y \: \: Used \: \: \text Integrals : } }}}$

$\maltese \: \: \: \displaystyle \rm\int \dfrac{1}{x} \: dx =log |x| +C$

$\maltese \: \: \: \displaystyle \rm\int {a}^{x} \: dx = \frac{ {a}^{x} }{log \: a} + C$

$\maltese \: \: \: \displaystyle \rm\int sinx \: dx = - cosx + C$

$\maltese \: \: \: \displaystyle \rm\int cosx \: dx =sinx + C$

$\maltese \: \: \: \displaystyle \rm\int sec {}^{2} x \: dx =tanx+ C$

$\maltese \: \: \: \displaystyle \rm\int {cosec}^{2} x \: dx = - cotx + C$

$\maltese \: \: \: \displaystyle \rm\int secx.tanx \: dx =secx + C$

$\maltese \: \: \: \displaystyle \rm\int cosecx.cotx \:dx = - cosecx + C$

Answer 2

Answer:

Step-by-step explanation:

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