In integral calculus, an irrational function $\dfrac{1}{(x-1)\sqrt{x+3}}$ is formed in $x$ and we have to evaluate the indefinite integration of this irrational function with respect to $x$.

$\displaystyle \int{\dfrac{1}{(x-1)\sqrt{x+3}}\,}dx$

In the denominator of the rational expression, one factor is a rational expression but the second factor is an irrational function. The combination pulls back us in evaluating the integral of the given function. So, we have to eliminate the irrationality of the second function.

It can be done by taking $v^2 \,=\, x+3$

Now, differentiate both sides of the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{\,(v^2)} \,=\, \dfrac{d}{dx}{\,(x+3)}$

$\implies$ $\dfrac{d}{dx}{\,(v^2)} \,=\, \dfrac{d}{dx}{\,(x)}+\dfrac{d}{dx}{\,(3)}$

$\implies$ $2v\dfrac{dv}{dx} \,=\, 1+0$

$\implies$ $2v\dfrac{dv}{dx} \,=\, 1$

$\implies$ $2v dv \,=\, 1 \times dx$

$\implies$ $2v dv \,=\, dx$

$\,\,\,\therefore\,\,\,\,\,\,$ $dx \,=\, 2vdv$

Now, express the mathematical expression $\displaystyle \int{\dfrac{1}{(x-1)\sqrt{x+3}}\,}dx$ in terms of $v$.

$(1) \,\,\,$ The differential eliminate $dx$ can be replaced by its equivalent value $2vdv$

$(2) \,\,\,$ The irrational function $\sqrt{x+3}$ can be replaced by $v$

But, there is a function $x-1$ still in terms of $x$. Now, we have to express it in $v$.

$(3) \,\,\,$ We have assumed that $v^2 \,=\, x+3$, then $x = v^2-3$. Now, $x-1$ $\,=\,$ $(v^2-3)-1$. Therefore, $x-1$ $\,=\,$ $v^2-4$

Now, let’s express the whole mathematical expression in $x$ into $v$

$\implies$ $\displaystyle \int{\dfrac{1}{(x-1)\sqrt{x+3}}\,}dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{(v^2-4)v}\,}(2vdv)$

The mathematical expression in $x$ is successfully transformed into $v$ and now, we have to simplify this expression.

$=\,\,\,$ $\displaystyle \int{\dfrac{1 \times 2v}{(v^2-4)v}\,}dv$

$=\,\,\,$ $\displaystyle \int{\dfrac{2v}{(v^2-4)v}\,}dv$

$=\,\,\,$ $\require{cancel} \displaystyle \int{\dfrac{2\cancel{v}}{(v^2-4)\cancel{v}}\,}dv$

$=\,\,\,$ $\displaystyle \int{\dfrac{2}{v^2-4}\,}dv$

$=\,\,\,$ $\displaystyle \int{\dfrac{2 \times 1}{v^2-4}\,}dv$

It can be simplified further by the constant multiple rule of integration.

$=\,\,\,$ $2 \times \displaystyle \int{\dfrac{1}{v^2-4}\,}dv$

$=\,\,\,$ $2 \times \displaystyle \int{\dfrac{1}{v^2-2^2}\,}dv$

The integration rule for the reciprocal of difference of squares can be used as a formula to evaluate the indefinite integral of the above function.

$=\,\,\,$ $2 \times \Bigg(\dfrac{1}{2 \times 2}\log_{e}{\Bigg|\dfrac{v-2}{v+2}\Bigg|}+c_1\Bigg)$

$=\,\,\,$ $\dfrac{2 \times 1}{2 \times 2}\log_{e}{\Bigg|\dfrac{v-2}{v+2}\Bigg|}+2 \times c_1$

$=\,\,\,$ $\dfrac{\cancel{2} \times 1}{\cancel{2} \times 2}\log_{e}{\Bigg|\dfrac{v-2}{v+2}\Bigg|}+2c_1$

$=\,\,\,$ $\dfrac{1 \times 1}{1 \times 2}\log_{e}{\Bigg|\dfrac{v-2}{v+2}\Bigg|}+c$

$=\,\,\,$ $\dfrac{1}{2}\log_{e}{\Bigg|\dfrac{v-2}{v+2}\Bigg|}+c$

Actually, the function is in $x$ but we have taken that $v = \sqrt{x+3}$. So, let’s express the expression in $v$ into $x$.

$=\,\,\,$ $\dfrac{1}{2}\log_{e}{\Bigg|\dfrac{\sqrt{x+3}-2}{\sqrt{x+3}+2}\Bigg|}+c$

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