We have, I=∫(2x 1)√(1 + x x^2) dx Let t=1+x x^2 dtdx=0+1 2x dt= (2x 1)dx Therefore, I= ∫√(t) dt I= t^3/23/2+C On putting the value ...

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

Integration by Partial Fractions

∫2x-1√1 + x -...

Question

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

Open in App

Solution

We have,
$I = \int (2 x - 1) \sqrt{1 + x - x^{2}} d x$

Let
$t = 1 + x - x^{2}$
$\frac{d t}{d x} = 0 + 1 - 2 x$
$d t = - (2 x - 1) d x$

Therefore,
$I = - \int \sqrt{t} d t$

$I = - \frac{t^{3 / 2}}{3 / 2} + C$

On putting the value of $t$ , we get
$I = - \frac{2}{3} (1 + x - x^{2})^{\frac{3}{2}} + C$

Hence, this is the answer.

Suggest Corrections

Similar questions

\int (x + 1) \sqrt{3 - 2 x - x^{2}} d x =

\int_{0}^{1} {tan}^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x

\int_{0}^{1} {tan}^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x

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

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

Join BYJU'S Learning Program