_ ^ x^2 + x - 6 (x - 2)(x - 1) dx =

MCQ

$\int_{}^{} {\frac{{{x^2} + x - 6}}{{(x - 2)(x - 1)}}dx = } $

A
$x + 2\log (x - 1) + c$
B
$2x + 2\log (x - 1) + c$
C
$x + 4\log (1 - x) + c$
✓
$x + 4\log (x - 1) + c$

✓

Answer

Correct option: D.

$x + 4\log (x - 1) + c$

d
(d)$\int_{}^{} {\frac{{{x^2} + x - 6}}{{(x - 2)(x - 1)}}\,dx} = \int_{}^{} {\frac{{(x + 3)(x - 2)}}{{(x - 2)(x - 1)}}\,dx} = \int_{}^{} {\frac{{x + 3}}{{x - 1}}\,dx} $
$ = \int_{}^{} {\frac{{x - 1}}{{x - 1}}\,dx + \int_{}^{} {\frac{4}{{x - 1}}\,dx} } = x + 4\log (x - 1) + c$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 7.1 inderfinite integral→7.1 inderfinite integral — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $\alpha$ and $\beta$ (where $\alpha > \beta$) are two zeroes of the equation $3\,cos{^{ - 1}}\left( {{x^2} - 5x - \frac{{11}}{2}} \right) = \pi $ then $(\alpha^2 + \beta^3)$ is -

→

A matrix consisting of a single column of m elements is know as:

Column matrix
Row matrix
Square matrix
Null matrix

→

The area bounded by the y-axis, $\text{y}=\cos\text{x}$ and $\text{y}=\sin\text{x}$ when $0\leq\text{x}\leq\frac{\pi}{2}$ is:

$2\big(\sqrt{2}-1\big)$
$\sqrt{2}-1$
$\sqrt{2}+1$
$\sqrt{2}$

→

For the $LP$ problem

Minimize $z=2 x+3 y$ the coordinates of the corner points of the bounded feasible region are $A\,(3,3), B\,(20,3),$ $\mathrm{C}\,(20,10), \mathrm{D}\,(18,12)$ and $\mathrm{E}\,(12,12) .$ The minimum value of $z$ is $\ldots \ldots$

→

If vectors $\vec A = 2i + 3j + 4k$, $\vec B = i + j + 5k$, and $\vec C$ form a left handed system, then $\vec C$ is

→

Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ be three unit vectors, such that $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=1$ and $\vec{\text{a}}$ is perpendicular to $\vec{\text{b}}.$ If $\vec{\text{c}}$ makes angle $\alpha$ and $\beta$ with $\vec{\text{a}}$ and $\vec{\text{b}}$ respectively, then $\cos\alpha+\cos\beta=$