Question

$\bf \: evaluate \:: \\ \displaystyle \rm = \int \frac{dx}{ 2 \sqrt{x}( \sqrt{x + {1}^{3} )} }$

Step by step explanation needed! ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle \rm \int \frac{dx}{ 2 \sqrt{x}( \sqrt{x + {1}^{3} )} }$

can be rewritten as

$\rm \:  =  \: \displaystyle \rm \int \frac{dx}{ 2 \sqrt{x}\sqrt{x +1} }$

can be further rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{2 \sqrt{ {x}^{2} + x } }$

Now, to evaluate this integral, we use method of Completing squares.

So, above integral can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{2 \sqrt{ {x}^{2} + x + \dfrac{1}{4} - \dfrac{1}{4} } }$

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{2 \sqrt{ {\bigg[x + \dfrac{1}{2} \bigg]}^{2} - {\bigg[\dfrac{1}{2} \bigg]}^{2} } }$

We know,

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log \bigg|x + \sqrt{ {x}^{2} - {a}^{2} } \bigg | + c}}$

So, using this, we get

$\rm \:  =  \: \dfrac{1}{2}log \bigg |x + \dfrac{1}{2} + \sqrt{ {\bigg[x + \dfrac{1}{2} \bigg]}^{2} - {\bigg[\dfrac{1}{2} \bigg]}^{2} } \bigg| + c$

$\rm \:  =  \: \dfrac{1}{2}log \bigg |x + \dfrac{1}{2} + \sqrt{ {x}^{2} + x} \bigg| + c$

$\rm \:  =  \: \dfrac{1}{2}log \bigg |\dfrac{2x + 1 + \sqrt{ {x}^{2} + x } }{2} \bigg| + c$

$\rm \:  =  \: \dfrac{1}{2}log \bigg |2x + 1 + \sqrt{ {x}^{2} + x } \bigg| - \dfrac{1}{2}log2 + c$

$\rm \:  =  \: \dfrac{1}{2}log \bigg |2x + 1 + \sqrt{ {x}^{2} + x } \bigg| + d \\ \\ \boxed{\tt{ where \: d = c - \frac{1}{2}log2 \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log \bigg|x + \sqrt{ {x}^{2} - {a}^{2} } \bigg | + c}}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log \bigg|x + \sqrt{ {x}^{2} + {a}^{2} } \bigg | + c}}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \dfrac{x}{a}+ c}}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ {x}^{2} + {a}^{2} } = \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c}}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a} log\bigg | \frac{x - a}{x + a} \bigg| + c}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle \rm \int \frac{dx}{ 2 \sqrt{x}( \sqrt{x + {1}^{3} )} }$

can be rewritten as

$\rm \:  =  \: \displaystyle \rm \int \frac{dx}{ 2 \sqrt{x}\sqrt{x +1} }$

can be further rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{2 \sqrt{ {x}^{2} + x } }$

Now, to evaluate this integral, we use method of Completing squares.

So, above integral can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{2 \sqrt{ {x}^{2} + x + \dfrac{1}{4} - \dfrac{1}{4} } }$

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{2 \sqrt{ {\bigg[x + \dfrac{1}{2} \bigg]}^{2} - {\bigg[\dfrac{1}{2} \bigg]}^{2} } }$

We know,

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log \bigg|x + \sqrt{ {x}^{2} - {a}^{2} } \bigg | + c}}$

So, using this, we get

$\rm \:  =  \: \dfrac{1}{2}log \bigg |x + \dfrac{1}{2} + \sqrt{ {\bigg[x + \dfrac{1}{2} \bigg]}^{2} - {\bigg[\dfrac{1}{2} \bigg]}^{2} } \bigg| + c$

$\rm \:  =  \: \dfrac{1}{2}log \bigg |x + \dfrac{1}{2} + \sqrt{ {x}^{2} + x} \bigg| + c$

$\rm \:  =  \: \dfrac{1}{2}log \bigg |\dfrac{2x + 1 + \sqrt{ {x}^{2} + x } }{2} \bigg| + c$

$\rm \:  =  \: \dfrac{1}{2}log \bigg |2x + 1 + \sqrt{ {x}^{2} + x } \bigg| - \dfrac{1}{2}log2 + c$

$\rm \:  =  \: \dfrac{1}{2}log \bigg |2x + 1 + \sqrt{ {x}^{2} + x } \bigg| + d \\ \\ \boxed{\tt{ where \: d = c - \frac{1}{2}log2 \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log \bigg|x + \sqrt{ {x}^{2} - {a}^{2} } \bigg | + c}}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log \bigg|x + \sqrt{ {x}^{2} + {a}^{2} } \bigg | + c}}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \dfrac{x}{a}+ c}}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ {x}^{2} + {a}^{2} } = \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c}}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a} log\bigg | \frac{x - a}{x + a} \bigg| + c}}$