CameraIcon

SearchIcon

MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Standard Formulae - 1

∫x3-x2+x-1x-1...

Question

$\int \frac{x^{3} - x^{2} + x - 1}{x - 1} d x$ is equal to
(where $C$ is constant of integration)

Open in App

Solution

$\int \frac{x^{3} - x^{2} + x - 1}{x - 1} d x = \int \frac{x^{2} (x - 1) + x - 1}{x - 1} d x = \int (x^{2} + 1) d x$
By taking the terms separately, we get
$= \int x^{2} d x + \int 1 d x$
Therefore, we obtain
$= \frac{x^{3}}{3} + x + C$

Suggest Corrections

thumbs-up

0

Similar questions

\int \frac{x + 1}{x^{\frac{1}{3}} + 1} d x

is equal to
(where

C

is constant of integration)

\int \frac{x^{3}}{(x + 1)^{2}} d x

is equal to
(where

C

is constant of integration)

x^{2} \neq n π + 1, n ϵ N

(the set of natural numbers), the integral

\int x \sqrt{\frac{2 sin (x^{2} - 1) - sin 2 (x^{2} - 1)}{2 sin (x^{2} - 1) + sin 2 (x^{2} - 1)}} d x

is equal to
(where

c

is a constant of integration).

Q. The value of the integral

\int \frac{e^{x^{2} + 4 ln x} - x^{3} e^{x^{2}}}{x - 1} d x

equal to
(where

C

is the constant of integration)

\int \frac{\sqrt{x}}{x + 1} d x = A \sqrt{x} + B {tan}^{- 1} \sqrt{x} + C,

| A | + | B |

is equal to
(where

C

is integration constant and

A

B

are fixed constants)

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Standard Formulae - 1

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon