Question

Find the integration of
$1 \div (9 {x}^{2} + 49)$
​

Answer 1

Given:

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{9x^2+49}\ dx}$

We have to get the integral into the form $\displaystyle\sf{k\int \dfrac{1}{u^2+1}\ du}$ because for a constant k and u is a function in x.

As we know ,

$\displaystyle\longrightarrow\sf{\int \dfrac{1}{u^2+1}\ du=\tan^{-1}(u)+c.}$

Solution :

" a " → Constant

Let us assume

$\displaystyle\longrightarrow\sf{u=ax\quad\iff\quad x=\dfrac{u}{a}}$

$\displaystyle\longrightarrow\sf{dx=\dfrac{1}{a}\ du}$

Now,

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{9x^2+49}\ dx=\int\dfrac{1}{9\left(\dfrac{u}{a}\right)^2+49}\cdot\dfrac{1}{a}\ du}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{9x^2+49}\ dx=\dfrac{1}{a}\int\dfrac{a^2}{9u^2+49a^2}\ du}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{9x^2+49}\ dx=\int\dfrac{a}{9\left(u^2+\dfrac{49a^2}{9}\right)}\ du}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{9x^2+49}\ dx=\dfrac{a}{9}\int\dfrac{1}{u^2+\dfrac{49a^2}{9}}\ du}$

Let's do comparison.

Comparing the RHS with $\displaystyle\sf{k\int\dfrac{1}{u^2+1}\ du,}$ we get,

$\displaystyle\longrightarrow\sf{\dfrac{49a^2}{9}=1}$

$\displaystyle\longrightarrow\sf{a=\pm\dfrac{3}{7}}$

Then,

$\displaystyle\longrightarrow\sf{k=\dfrac{a}{9}=\pm\dfrac{1}{21}}$

$\displaystyle\longrightarrow\sf{u=\pm\dfrac{3}{7}\ x}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{9x^2+49}\ dx=\pm\dfrac{1}{21}\int\dfrac{1}{u^2+1}\ du}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{9x^2+49}\ dx=\pm\dfrac{1}{21}\tan^{-1}(u)+c}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{9x^2+49}\ dx=\pm\dfrac{1}{21}\tan^{-1}\left(\pm\dfrac{3}{7}\ x\right)+c}$

So, the final answer

$\displaystyle\longrightarrow\sf{\underline{\underline{\int\dfrac{1}{9x^2+49}\ dx=\dfrac{1}{21}\tan^{-1}\left(\dfrac{3}{7}\ x\right)+c}}}$

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Integrals :-

$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}$