Question

3rd one plz solve it ​

Answer 1

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int \dfrac{1}{4 + 9 {x}^{2} } \: dx$

Can be written as:

$\displaystyle \rm \longrightarrow I = \int \dfrac{1}{9( {x}^{2} + \frac{4}{9} )} \: dx$

$\displaystyle \rm \longrightarrow I = \int \dfrac{1}{9} \cdot \dfrac{1}{{x}^{2} + \frac{4}{9}} \: dx$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{9} \int \dfrac{1}{{x}^{2} + \frac{4}{9}} \: dx$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{9} \int \dfrac{1}{{x}^{2} + ( \frac{2}{3})^{2} } \: dx$

We know that:

$\displaystyle \rm \longrightarrow \int \dfrac{1}{{x}^{2} + {a}^{2} } \: dx = \dfrac{1}{a} \cdot \tan^{ - 1} \bigg ( \frac{x}{a} \bigg ) + C$

So, the integral becomes:

$\displaystyle \rm \longrightarrow I = \dfrac{1}{9} \times \dfrac{1}{ \frac{2}{3} } \times \tan^{ - 1} \bigg( \dfrac{x}{ \frac{2}{3} } \bigg) +C$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{9} \times \dfrac{3}{2} \times \tan^{ - 1} \bigg( \dfrac{3x}{2} \bigg) +C$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{6} \tan^{ - 1} \bigg( \dfrac{3x}{2} \bigg) +C$

Therefore:

$\displaystyle \rm \longrightarrow \int \dfrac{1}{4 + 9 {x}^{2} } \: dx = \dfrac{1}{6} \tan^{ - 1} \bigg( \dfrac{3x}{2} \bigg) +C$

★ Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$