Question
Evalute : $\int \frac{3 x-1}{2 x^2-x-1} d x$

Answer

$
\text { Let } \begin{aligned}
I & =\int \frac{3 x-1}{2 x^2-x-1} d x \\
& =\int \frac{3 x-1}{(x-1)(2 x+1)} d x
\end{aligned}
$
Let $\frac{3 x-1}{(x-1)(2 x+1)}=\frac{A}{x-1}+\frac{B}{2 x+1}$
$
\therefore 3 x-1=A(2 x+1)+B(x-1)
$
Put $x-1=0$, i.e. $x=1$, we get
$
\begin{aligned}
& 3(1)-1=A(3)+B(0) \quad \therefore 2=3 A \quad \therefore A=\frac{2}{3} \\
\end{aligned}
$
Put $2 x+1=0$, i.e. $x=-\frac{1}{2}$, we get
$
\begin{aligned}
& 3\left(-\frac{1}{2}\right)-1=A(0)+B\left(-\frac{3}{2}\right) \\
& \therefore-\frac{5}{2}=-\frac{3}{2} B \quad \therefore B=\frac{5}{3} \\
& \therefore \frac{3 x-1}{(x-1)(2 x+1)}=\frac{\left(\frac{2}{3}\right)}{x-1}+\frac{\left(\frac{5}{3}\right)}{2 x+1} \\
& \therefore I=\int\left[\frac{\left(\frac{2}{3}\right)}{x-1}+\frac{\left(\frac{5}{3}\right)}{2 x+1}\right] d x \\
& \quad=\frac{2}{3} \int \frac{1}{x-1} d x+\frac{5}{3} \int \frac{1}{2 x+1} d x \\
& \quad=\frac{2}{3} \log |x-1|+\frac{5}{3} \cdot \frac{\log |2 x+1|}{2}+c \\
& \quad=\frac{2}{3} \log |x-1|+\frac{5}{6} \log |2 x+1|+c .
\end{aligned}
$

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

Similar questions

Calculate Walsh’s Price Index Number.
CommodityBase YearCurrent Year
PriceQuantityPriceQuantity
I1016209
II202258
III3034027
IV6097536
A firm manufactures two products $A$ and $B$ on which profit is earned per unit $₹ 3 /-$ and $₹ 4 /$ - respectively. Each product is processed on two machines $M _1$ and $M _2$. Product A requires one minute of processing time on Mx and two minutes of processing time on $M _2$. B requires one minute of processing time on $M _1$ and one minute processing time on $M_2$ Machine $M_1$ is available for use for $450$ minutes while $M_2$ is available for $600$ minutes during any working day. Final the number of units of products $A$ and $B$ to be manufactured to get the maximum profit.
In each of the following examples, verify that the given function is a solution of the corresponding differential equation:
SolutionD.E.
(i)xy=log y+ky^(')(1-xy)=y^(2)
(ii)y=x^(n)x^(2)(d^(2)y)/(dx^(2))-nx(dy)/(dx)+ny=0
(iii)y=e^(x)(dy)/(dx)=y
(iv)y=1-log xx^(2)(d^(2)y)/(dx^(2))=1
(v)y=ae^(x)+be^(-x)(d^(2)y)/(dx^(2))=y
(vi)ax^(2)+by^(2)=5xy(d^(2)y)/(dx^(2))+x((dy)/(dx))^(2)=y*(dy)/(dx)
Minimize z = 4x + 2y, Subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0, y ≥ 0
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.

$f(x)= \begin{cases}k\left(4-x^2\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}$

Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)

If $x ^7 \cdot y ^9=( x + y )^{16}$, then show that $\frac{d y}{d x}=\frac{y}{x}$.
Solve the following equations by the method of inversion : $2x – y + z = 1, x + 2y + 3z = 8$ and $3x + y – 4z = 1$
The following table shows the amount of sugar production (in lac tonnes) for the years 1971 to 1982.
YEARPRODUCTIONYEARPRODUCTION
1971119773
1972019786
1973119795
1974219801
1975319814
19762198210
Fit a trend line to the above data by graphical method.
Write the negations of the following statements:
(i) 7 is a prime number and the Taj Mahal is in Agra.
(ii) 10 > 5 and 3 < 8.
(iii) I will have tea or coffee.
(iv) ∀n ∈ N, n + 3 > 9.
(v) ∃x ∈ A, such that x + 5 < 11.
$\int_1^2 \frac{5 x^2}{x^2+4 x+3} d x$