Indefinite IntegrationmediumFree

Trace of AB then Integrate 3/f(x) | JEE

JEE Maths question with a full step-by-step solution.

Question

Let

A=\begin{bmatrix} x^{2}&6&0\\ 1&-5&1\\ 2&0&x\end{bmatrix}

and

B=\begin{bmatrix} 4&0&0\\ 0&1&0\\ 0&0&8\end{bmatrix}

. If

f(x)=\operatorname{trace}(AB)

, then

\displaystyle\int \dfrac{3\,dx}{f(x)}

\dfrac{1}{4}\ln\left|\dfrac{2x-1}{2x+5}\right|+c

correct

\dfrac{1}{4}\ln\left|\dfrac{2x+5}{2x-1}\right|+c

\dfrac{1}{3}\ln\left|\dfrac{1-2x}{2x+5}\right|+c

\dfrac{1}{3}\ln\left|\dfrac{1-2x}{2x+3}\right|+c

Solution

Step 1: Compute the trace of

AB

. Since

B=\operatorname{diag}(4,1,8)

scales column

j

A

by the

j

th diagonal entry, the diagonal entries of

AB

are

(AB)_{11}=4x^{2},\qquad (AB)_{22}=-5,\qquad (AB)_{33}=8x

\therefore f(x)=\operatorname{trace}(AB)=4x^{2}-5+8x=4x^{2}+8x-5

Step 2: Set up the integral and pull out the leading coefficient:

\int\dfrac{3\,dx}{4x^{2}+8x-5}=\dfrac{3}{4}\int\dfrac{dx}{x^{2}+2x-\tfrac{5}{4}}

Step 3: Complete the square in the denominator:

x^{2}+2x-\dfrac{5}{4}=(x+1)^{2}-1-\dfrac{5}{4}=(x+1)^{2}-\dfrac{9}{4}=(x+1)^{2}-\left(\dfrac{3}{2}\right)^{2}

Step 4: Apply

\displaystyle\int\dfrac{du}{u^{2}-a^{2}}=\dfrac{1}{2a}\ln\left|\dfrac{u-a}{u+a}\right|

with

u=x+1,\ a=\dfrac{3}{2}

=\dfrac{3}{4}\cdot\dfrac{1}{2\cdot\tfrac{3}{2}}\ln\left|\dfrac{(x+1)-\tfrac{3}{2}}{(x+1)+\tfrac{3}{2}}\right|+c=\dfrac{3}{4}\cdot\dfrac{1}{3}\ln\left|\dfrac{x-\tfrac{1}{2}}{x+\tfrac{5}{2}}\right|+c=\dfrac{1}{4}\ln\left|\dfrac{2x-1}{2x+5}\right|+c

Correct answer: (1)

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