Evalute : x (x-1)^2(x+2) d x - Vidyadip

Question

Evalute : $\int \frac{x}{(x-1)^2(x+2)} d x$

✓

Answer

Let $I =\int \frac{x}{(x-1)^2(x+2)} d x$
Let $\frac{x}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}$
$
\therefore x = A ( x -1)( x +2)+ B ( x +2)+ C ( x -1)^2
$
Put $x-1=0$, i.e. $x=1$, we get
$
\begin{aligned}
& 1= A (0)(3)+ B (3)+ C (0) \\
& \therefore B =\frac{1}{3}
\end{aligned}
$
Put $x+2=0$, i.e. $x=-2$, we get
$
\begin{aligned}
& -2= A (-3)(0)+ B (0)+ C (9) \\
& \therefore C =-\frac{2}{9}
\end{aligned}
$
Put $x =-1$, we get,
$
-1=A(-2)(1)+B(1)+C(4)
$
But $B =\frac{1}{3}$ and $C =-\frac{2}{9}$
$
\begin{aligned}
& \therefore-1=-2 A+\frac{1}{3}-\frac{8}{9} \\
& \therefore 2 A=-\frac{5}{9}+1=\frac{4}{9} \quad \therefore A=\frac{2}{9}
\end{aligned}
$
$
\begin{aligned}
& \therefore \frac{x}{(x-1)^2(x+2)}=\frac{(2 / 9)}{x-1}+\frac{(1 / 3)}{(x-1)^2}+\frac{(-2 / 9)}{x+2} \\
& \therefore I=\int\left[\frac{(2 / 9)}{x-1}+\frac{(1 / 3)}{(x-1)^2}+\frac{(-2 / 9)}{x+2}\right] d x \\
& \quad=\frac{2}{9} \int \frac{1}{x-1} d x+\frac{1}{3} \int(x-1)^{-2} d x-\frac{2}{9} \int \frac{1}{x+2} d x
\end{aligned}
$
$
=\frac{2}{9} \log |x-1|+\frac{1}{3} \cdot \frac{(x-1)^{-1}}{-1}-\frac{2}{9} \log |x+2|+c
$
$
=\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+c \text {. }
$

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 "Solve the Following Question.(5 Marks)" from Integration (p-1)→Integration (p-1) — all question types→Maths (commerce) STD 12 Commerce / Arts question papers→Maharashtra Board STD 12 Commerce / Arts question papers→Maharashtra Board question paper generator→

Similar questions

Find inverse of the following matrices (if they exist) by elementary transformation : $\left[\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$

A manufacturer makes a clear profit of 30% on the cost after allowing a 35% discount. If the cost of production rises by 20%, by what percentage should he reduce the rate of discount so as to make the same rate of profit keeping his list prices unaltered.

The total cost of $3 \ T.V$. and $2\ \text{V.C.R}$. is $₹\ 35,000$. The shopkeeper wants profit of $₹\ 1000$ per television and $₹ \ 500$ per $\text{V.C.R}.$ He can sell $2\ T.V$. and $1\ \text{ V.C.R}$. and get the total revenue as $₹\ 21,500$. Find the cost price and the selling price of a $T.V$. and a $\text{V.C.R}$.

The sum of three numbers is $6$. If we multiply the third number by $3$ and add it to the second number, we get $11$. By adding first and third numbers we get a number that is double the second number. Use this information and find a system of linear equations. Find the three numbers using matrices.

Calculate Walsh’s Price Index Number.

Commodity	Base Year		Current Year
	Price	Quantity	Price	Quantity
I	10	16	20	9
II	20	2	25	8
III	30	3	40	27
IV	60	9	75	36

Evalute : $\int \frac{3 x-2}{(x+1)^2(x+3)} d x$

Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.

Minimize z = 2x + 3y, Subject to x – y ≤ 1, x + y ≥ 3, x ≥ 0, y ≥ 0

A chemical company produces a chemical containing three basic elements A, B, C so that it has at least 16 liters of A, 24 liters of B, and 18 liters of C. This chemical is made by mixing two compounds I and II. Each unit of compound I has 4 liters of A, 12 liters of B, 2 liters of C. Each unit of compound II has 2 liters of A, 2 liters of B and 6 liters of C. The cost per unit of compound is ₹ 800/- and that of compound II is ₹ 640/- Formulate the problem as L.P.P and solve it to minimize the cost.

If $A=\left[\begin{array}{cc}2 & -4 \\ 3 & -2 \\ 0 & 1\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 2 \\ -2 & 1 & 0\end{array}\right]$ then showthat $(A B)^{\top}=B^{\top} A^{\top}$.