Question

integration✌️


don't post irrelevant answers

Answer 1

Given,

$\displaystyle\longrightarrow I=\int\dfrac{x^2-1}{x^4-1}\ dx$

Factorising the denominator,

$\displaystyle\longrightarrow I=\int\dfrac{x^2-1}{(x^2-1)(x^2+1)}\ dx$

Cancelling like terms,

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

$\displaystyle\longrightarrow\underline{\underline{I=\tan^{-1}x+C}}$

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Given,

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

Factoring the denominator,

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

$\displaystyle\longrightarrow I=\int\dfrac{x^2+1}{(x^2+1)^2-(x\sqrt2)^2}\ dx}$

$\displaystyle\longrightarrow I=\int\dfrac{x^2+1}{(x^2+x\sqrt2+1)(x^2-x\sqrt2+1)}\ dx}$

Multiplying and dividing by 2,

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{2x^2+2}{(x^2+x\sqrt2+1)(x^2-x\sqrt2+1)}\ dx}$

Adding and subtracting $x\sqrt2$ in the numerator,

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{2x^2+x\sqrt2-x\sqrt2+2}{(x^2+x\sqrt2+1)(x^2-x\sqrt2+1)}\ dx}$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{(x^2-x\sqrt2+1)+(x^2+x\sqrt2+1)}{(x^2+x\sqrt2+1)(x^2-x\sqrt2+1)}\ dx}$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\left(\dfrac{1}{x^2+x\sqrt2+1}+\dfrac{1}{x^2-x\sqrt2+1}\right)\ dx}$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{1}{x^2+x\sqrt2+1}\ dx+\dfrac{1}{2}\int\dfrac{1}{x^2-x\sqrt2+1}\ dx}\quad\quad\dots(1)$

Now let us evaluate each the integral.

$\displaystyle\longrightarrow \int\dfrac{1}{x^2\pm x\sqrt2+1}\ dx=\int\dfrac{1}{x^2\pm x\sqrt2+\frac{1}{2}+\frac{1}{2}}\ dx$

$\displaystyle\longrightarrow \int\dfrac{1}{x^2\pm x\sqrt2+1}\ dx=\int\dfrac{1}{\left(x\pm\frac{1}{\sqrt2}\right)^2+\frac{1}{2}}\ dx$

Taking $\dfrac{1}{2}$ common from denominator,

$\displaystyle\longrightarrow \int\dfrac{1}{x^2\pm x\sqrt2+1}\ dx=\int\dfrac{1}{\frac{1}{2}\left(x\sqrt2\pm1\right)^2+\frac{1}{2}}\ dx$

$\displaystyle\longrightarrow \int\dfrac{1}{x^2\pm x\sqrt2+1}\ dx=2\int\dfrac{1}{\left(x\sqrt2\pm1\right)^2+1}\ dx$

Substitute,

$\longrightarrow u=x\sqrt2\pm1$

$\longrightarrow dx=\dfrac{1}{\sqrt2}\ du$

Then,

$\displaystyle\longrightarrow \int\dfrac{1}{x^2\pm x\sqrt2+1}\ dx=\sqrt2\int\dfrac{1}{u^2+1}\ du$

$\displaystyle\longrightarrow \int\dfrac{1}{x^2\pm x\sqrt2+1}\ dx=\sqrt2\tan^{-1}u$

$\displaystyle\longrightarrow \int\dfrac{1}{x^2\pm x\sqrt2+1}\ dx=\sqrt2\tan^{-1}(x\sqrt2\pm1)$

Thus (1) becomes,

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{\sqrt2}\tan^{-1}\left(x\sqrt2+1\right)+\dfrac{1}{\sqrt2}\tan^{-1}\left(x\sqrt2-1\right)+C}}$

Answer 2

Answer:

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