If ∫1/(x^2+4x^2+9)dx=Atan^(-1)⁡x/2+Btan^(-1)⁡x/3+C then A-B=

Correct answer: (A) 1/6. Since 1/AB=1/(B-A)1/A-1/B ⇒ ∫1/(x^2+4x^2+9)dx= ∫1/51/(x^2+4)-1/(x^2+9)dx =1/51/2tan^(-1)⁡x/2-1/3tan^(-1)⁡x/3+C ⇒ A=1/10 B=-1/15 ⇒A−B=1/10 +1/15 =5/30 =1/6

Q76 / 150 · MHT CET 2017 · Mathematics

If $\int \frac{1}{(x^{2} + 4) (x^{2} + 9)} d x = A \tan^{- 1} \frac{x}{2} + B \tan^{- 1} (\frac{x}{3}) + C$ then $A - B =$

Correct answer
A
$\frac{1}{6}$
B
$\frac{1}{30}$
C
$- \frac{1}{30}$
D
$- \frac{1}{6}$

Solution

Since

\frac{1}{A B} = \frac{1}{B - A} (\frac{1}{A} - \frac{1}{B})

\Rightarrow

\int \frac{1}{(x^{2} + 4) (x^{2} + 9)} d x = \int \frac{1}{5} (\frac{1}{x^{2} + 4} - \frac{1}{x^{2} + 9}) d x

= \frac{1}{5} [\frac{1}{2} \tan^{- 1} \frac{x}{2} - \frac{1}{3} \tan^{- 1} \frac{x}{3}] + C

\Rightarrow

A = \frac{1}{10} B = - \frac{1}{15}

\Rightarrow A - B = \frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6}

Hence, the correct answer is option: 'A'.

If ∫1x2+4x2+9dx=Atan-1⁡x2+Btan-1⁡x3+C then A-B=

Solution

Relevant Questions

If $\int \frac{1}{(x^{2} + 4) (x^{2} + 9)} d x = A \tan^{- 1} \frac{x}{2} + B \tan^{- 1} (\frac{x}{3}) + C$ then $A - B =$