Integrate the following with respect to $x$: (i) $\dfrac{2x-3}{x^2+4x-12}$ (ii) $\dfrac{5x-2}{2+2x+x^2}$ (iii) $\dfrac{3x+1}{2x^2-2x+3}$

Integrate the following with respect to x: (i) (2x-3)/(x²+4x-12) (ii) (5x-2)/(2+2x+x²) (iii) (3x+1)/(2x²-2x+3)

Exercise 11.11 · Q1

x

(i)

\dfrac{2x-3}{x^2+4x-12}

(ii)

\dfrac{5x-2}{2+2x+x^2}

(iii)

\dfrac{3x+1}{2x^2-2x+3}

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★

55% · 71/129 Questions

✓ Free question

Concept understanding — Integration of Rational Algebraic Functions

Several rational and irrational integrands reduce to a small table of standard forms.

Type I — pure quadratic denominators. With constant $a>0$ :

\int\frac{dx}{a^2-x^2}=\frac1{2a}\log\left|\frac{a+x}{a-x}\right|+c,\quad \int\frac{dx}{x^2-a^2}=\frac1{2a}\log\left|\frac{x-a}{x+a}\right|+c,\quad \int\frac{dx}{a^2+x^2}=\frac1a\tan^{-1}\!\frac{x}{a}+c,

\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\!\frac{x}{a}+c,\quad \int\frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+c,\quad \int\frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+c.

Type II — general quadratic denominators $\int\frac{dx}{ax^2+bx+c}$ and $\int\frac{dx}{\sqrt{ax^2+bx+c}}$ : make the coefficient of $x^2$ unity, complete the square to write $ax^2+bx+c$ as a sum/difference of two squares, and reduce to Type I.

Type III — linear over quadratic $\int\frac{px+q}{ax^2+bx+c}\,dx$ (and the $\sqrt{\ }$ version): write $px+q=A\frac{d}{dx}(ax^2+bx+c)+B$ , find $A,B$ by matching coefficients, then split into a log/derivative-over-function piece plus a Type-I/II piece.

Type IV — square roots of quadratics (Result 11.3), via integration by parts:

\int\sqrt{a^2-x^2}\,dx=\frac x2\sqrt{a^2-x^2}+\frac{a^2}2\sin^{-1}\!\frac xa+c,

\int\sqrt{x^2-a^2}\,dx=\frac x2\sqrt{x^2-a^2}-\frac{a^2}2\log\left|x+\sqrt{x^2-a^2}\right|+c,\quad \int\sqrt{x^2+a^2}\,dx=\frac x2\sqrt{x^2+a^2}+\frac{a^2}2\log\left|x+\sqrt{x^2+a^2}\right|+c.

Tip

Match the sign pattern first — $a^2-x^2$ leads to $\sin^{-1}$ , $a^2+x^2$ to $\tan^{-1}$ or a $\log(x+\sqrt{\ })$ , and $x^2-a^2$ to a $\log$ — before completing the square or splitting the numerator.

Write the numerator as $A\cdot(\text{derivative of the denominator})+B$ , split the integral into a log piece and a Type I/II piece.

(i) Let $I=\displaystyle\int\frac{2x-3}{x^2+4x-12}\,dx$ .

Step 1. Write $2x-3=A(2x+4)+B$ . Comparing coefficients: $2A=2\Rightarrow A=1$ ; $4A+B=-3\Rightarrow B=-7$ .

Step 2. $I=\displaystyle\int\frac{2x+4}{x^2+4x-12}\,dx-7\int\frac{dx}{x^2+4x-12}=\log|x^2+4x-12|-7\int\frac{dx}{(x+2)^2-16}$ .

Step 3. $\displaystyle\int\frac{dx}{(x+2)^2-16}=\dfrac1{8}\log\left|\dfrac{x-2}{x+6}\right|+c$ (Type I with $u=x+2$ , $a=4$ ). So $I=\log|x^2+4x-12|-\dfrac78\log\left|\dfrac{x-2}{x+6}\right|+c$ .

(ii) Let $I=\displaystyle\int\frac{5x-2}{x^2+2x+2}\,dx$ .

Step 1. Write $5x-2=A(2x+2)+B$ . $2A=5\Rightarrow A=\dfrac52$ ; $2A+B=-2\Rightarrow B=-7$ .

Step 2. $I=\dfrac52\displaystyle\int\frac{2x+2}{x^2+2x+2}\,dx-7\int\frac{dx}{(x+1)^2+1}=\dfrac52\log|x^2+2x+2|-7\tan^{-1}(x+1)+c$ .

(iii) Let $I=\displaystyle\int\frac{3x+1}{2x^2-2x+3}\,dx$ .

Step 1. Write $3x+1=A(4x-2)+B$ . $4A=3\Rightarrow A=\dfrac34$ ; $-2A+B=1\Rightarrow B=\dfrac52$ .

Step 2. $I=\dfrac34\displaystyle\int\frac{4x-2}{2x^2-2x+3}\,dx+\dfrac52\int\frac{dx}{2x^2-2x+3}=\dfrac34\log|2x^2-2x+3|+\dfrac54\int\frac{dx}{x^2-x+\tfrac32}$ .

Step 3. $x^2-x+\tfrac32=\left(x-\tfrac12\right)^2+\tfrac54$ , a Type I form with $a=\tfrac{\sqrt5}2$ : $\dfrac54\int\frac{dx}{(x-\frac12)^2+(\frac{\sqrt5}2)^2}=\dfrac54\cdot\dfrac2{\sqrt5}\tan^{-1}\!\left(\dfrac{2x-1}{\sqrt5}\right)=\dfrac{\sqrt5}2\tan^{-1}\!\left(\dfrac{2x-1}{\sqrt5}\right)$ .

✓Final answer

\log\left|x^2+4x-12\right|-\dfrac78\log\left|\dfrac{x-2}{x+6}\right|+c

←Previous question · 11.7.9Q3 — Find the integrals of the following: (i) dfrac1√((2+x)²-1) (ii) dfrac1…→Next question · 11.7.9Q2 — Integrate the following with respect to x: (i) dfrac2x+1√(9+4x-x²) (ii…

Q.Integrate the following with respect to xxx:

🎓Unlock everything free for 14 days

Q.Integrate the following with respect to $x$ :

Unlock everything free for 14 days