The value of -0-22x-3 - 6x-2 - 9x - 5-x-2 - 2x - 5dx-equals4-1None of the above1

The correct option is C None of the aboveMake the substitution x – 1= t. ∫_ 1^12t+1^3 6t+1^2 + 9t+1 5/t+1^2 2t+1 + 5dt ∫_ 1^12t^3 +3t/t^2+4dt = 0 Since, ...

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

Byju's Answer

Standard XII

Mathematics

Area between Two Curves

The value of ...

Question

The value of $\int_{0}^{2} \frac{2 x^{3} - 6 x^{2} + 9 x - 5}{x^{2} - 2 x + 5} d x$ equals

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

-1

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

None of the above

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

Open in App

Solution

The correct option is D None of the above
Make the substitution x – 1= t.
$\int_{- 1}^{1} \frac{2 {t + 1}^{3} - 6 {t + 1}^{2} + 9 t + 1 - 5}{{t + 1}^{2} - 2 t + 1 + 5} d t$
$\int_{- 1}^{1} \frac{2 t^{3} + 3 t}{t^{2} + 4} d t$
= 0
Since, the above expression is an odd function. And limits are equal and opposite in sign.

Suggest Corrections

Similar questions

Q. Evaluate:

\int \frac{x^{2} + 4 x}{x^{3} + 6 x^{2} + 5} d x

Find the quotient and remainder when each of the polynomials below is divided by 2x + 1.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Q. The value of

\int_{0}^{2} \frac{2 x^{3} - 6 x^{2} + 9 x - 5}{x^{2} - 2 x + 5} d x

equals

$\begin{matrix} p x - q x^{2} + 2 x + 1 3 x^{3} + x^{2} + 2 x + 6 3 x^{3} + 6 x^{2} + 3 x - -------------- - - 5 x^{2} - x + 5 - 5 x^{2} - 10 x - 5 - --------------- - 9 x + 10 \end{matrix}$

Q. Evaluate:

\int \sqrt{x^{2} + 2 x + 5} d x

Join BYJU'S Learning Program

The value of ∫202x3−6x2+9x−5x2−2x+5dx equals

The correct option is D None of the aboveMake the substitution x – 1= t. ∫1−12t+13−6t+12+9t+1−5t+12−2t+1+5dt ∫1−12t3+3tt2+4dt = 0 Since, the above expression is an odd function. And limits are equal and opposite in sign.

The value of $\int_{0}^{2} \frac{2 x^{3} - 6 x^{2} + 9 x - 5}{x^{2} - 2 x + 5} d x$ equals