Question

integration of 1/9x'2+49 dx​

Answer 1

Question:-

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

Answer:-

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

Solution:-

Given,

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

Our aim is to get the integral to the form $\displaystyle\sf{k\int \dfrac{1}{u^2+1}\ du}$ for a constant k and u is a function in x, because we know that,

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

Let,

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

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

where 'a' is a constant. Then,

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

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

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

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

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

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

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

And,

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

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

Then,

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

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

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

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

Answer 2

Before solving this problem, let us know a formula:

$\displaystyle\mathrm{\quad \int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\:tan^{-1}\big(\frac{x}{a}\big)+c}$

where c is constant of integration.

Now we proceed to solve the problem:

$\displaystyle\therefore \mathrm{\int\frac{dx}{9x^{2}+49}}$

$\displaystyle\mathrm{=\int \dfrac{dx}{(3x)^{2}+7^{2}}}$

Let $\displaystyle\mathrm{3x=z\implies dx=\frac{1}{3}\:dz}$

Then we have $\displaystyle\to$

$\displaystyle\mathrm{=\frac{1}{3} \int \frac{dz}{z^{2}+7^{2}}}$

$\displaystyle\mathrm{=\frac{1}{3} \times \frac{1}{7}\:tan^{-1}\big(\frac{z}{7}\big)+c}$

$\quad$ where c is constant of integration

$\displaystyle\mathrm{=\frac{1}{21}\:tan^{-1}\big(\frac{3x}{7}\big)+c}$

$\quad\quad$ since $\displaystyle\mathrm{z=3x}$

$\displaystyle\to \boxed{\mathrm{\int\dfrac{dx}{9x^{2}+49}=\frac{1}{21}\:tan^{-1}\big(\frac{3x}{7}\big)+c}}$

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