Question

Find the integral of x^2-2x+1
please answer ......​

Answer 1

Answer:

answer is 1 very simple

Answer 2

ANSWER:

$\sf \displaystyle \int (x^2-2x+1)dx=\left(\dfrac{2}{3}\right)x^{3}-x^{2}+x + C$

GIVEN:

• x² - 2x + 1

TO FIND:

• The integral of x² - 2x + 1.

EXPLANATION:

$\sf \displaystyle \int (x^2-2x+1)dx$

$\sf \displaystyle \int x^2\ dx- \int2x\ dx+ 1\int\ dx$

$\boxed{\bold{\large{\gray{\int x^n = \dfrac{x^{n+1}}{n+1}}}}}$

$\sf 2\left(\dfrac{x^{2+1}}{2+1}\right)-2\left(\dfrac{x^{1+1}}{1+1}\right)+\left(\dfrac{x^{0+1}}{0+1}\right)+C$

$\sf 2\left(\dfrac{x^{3}}{3}\right)-2\left(\dfrac{x^{2}}{2}\right)+\left(\dfrac{x^{1}}{1}\right) + C$

$\sf \left(\dfrac{2}{3}\right)x^{3}-\left(\dfrac{2}{2}\right)x^{2}+x + C$

$\sf \left(\dfrac{2}{3}\right)x^{3}-x^{2}+x + C$

Here C is the constant of integration.

OTHER FORMULAE:

$\boxed{\begin{minipage}{7 cm}\int sin\ x \ dx=- cos\ x + C\\ \\\int cos\ x \ dx=sin\ x + C\\ \\ \int sec^2\ x \ dx=tan\ x + C\\ \\\int cosec^2\ x\ dx =-cot\ x + C\\ \\ \int sec\ x \ tan \ x\ dx=sec\ x + C\\ \\ \int cosec\ x \ cot \ x\ dx=-cosec\ x + C\\ \\ \int \dfrac{1}{\sqrt{1-x^2}}dx=sin^{-1}\ x + C\\ \\ \int \dfrac{1}{\sqrt{1-x^2}}dx=-cos^{-1}\ x + C\\ \\ \int \dfrac{1}{1+x^2}dx=tan^{-1}\ x + C \end{minipage}}$