The correct option is A x^22 x+C I=∫( x 1 ) dx =∫ x dx ∫ dx I= x ^ 2 2 x+C {∵∫ x ^ n dx = x ^ n+1 n+1 +C} ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Theorems on Integration

Evaluate: ∫...

Question

Evaluate: $\int (x - 1) d x$

\frac{x^{2}}{2} - x + C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{x^{2}}{2} - 2 x + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{x^{2}}{2} - 1 + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{x^{2}}{2} - 2 + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{x^{2}}{2} - x + C$
$I = \int (x - 1) d x = \int x d x - \int d x I = \frac{x^{2}}{2} - x + C {∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C}$

Suggest Corrections

Similar questions

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

\int \frac{(1 - x) (2 + x)}{x} d x =

\int \frac{x^{2} - 1}{(x^{2} + 1) \sqrt{x^{4} + 1}} d x = {sec}^{- 1} ∣ ∣ ∣ \frac{x^{2} + 1}{x \sqrt{2}} ∣ ∣ ∣ + C

\int \frac{d t}{t \sqrt{t^{2} - a}} = \frac{1}{\sqrt{a}} {sec}^{- 1} ∣ ∣ ∣ \frac{t}{\sqrt{a}} ∣ ∣ ∣ + C

I = \int \frac{d x}{1 + \sqrt{x^{2} + 2 x + 2}} = \frac{A}{7} log ∣ ∣ x + 1 + \sqrt{x^{2} + 2 x + 2} ∣ ∣ - \frac{\sqrt{x^{2} + 2 x + 2} - 1}{x + 1} + C

\int_{- 1}^{1} \frac{x + | x | + 1}{x^{2} + 2 | x | + 1} d x .

Join BYJU'S Learning Program