Question

1/√2ˣ²+3 + 1/7ˣ²+3,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

Hello,

Answer:
$= log |x + \sqrt{ \frac{2 {x}^{2} + 3 }{2} } | + \frac{1}{ \sqrt{21} } {tan}^{ - 1} ( \frac{x \sqrt{7} }{ \sqrt{3} } ) + c$

Solution:

∫ $\frac{1}{ \sqrt{2 {x}^{2} + 3 } } + \frac{1}{7 {x}^{2} + 3 } dx$
Convert both terms into 1/
${x}^{2} + {a}^{2}$

form
$=∫ \frac{1}{ \sqrt{2} \sqrt{ {x}^{2} + ( { \sqrt{ \frac{3}{2} } })^{2} } } + \frac{1}{7( {x}^{2} + { \sqrt{ \frac{3}{7} } })^{2} } dx$

$= \frac{1}{ \sqrt{2} } log( |x + \sqrt{( {x}^{2} + ( { \sqrt{ \frac{3}{2} } })^{2} } | ) ) + \frac{ \sqrt{7} }{7 \sqrt{3} } {tan}^{ - 1} ( \frac{x \sqrt{7} }{ \sqrt{3} } + c$
is the answer,you can simplify it more if possible.

like
$= log |x + \sqrt{ \frac{2 {x}^{2} + 3 }{2} } | + \frac{1}{ \sqrt{21} } {tan}^{ - 1} ( \frac{x \sqrt{7} }{ \sqrt{3} } ) + c$
Hope it helps you.

Answer 2

I guessed your question is $\int{\frac{1}{\sqrt{2x^2+3}}+\frac{1}{7x^2+3}}\,dx$


=$\int{\frac{1}{\sqrt{2}}\frac{1}{\sqrt{x^2+\frac{3}{2}}}+\frac{1}{7}\frac{1}{x^2+\frac{3}{7}}}\,dx$


=$\frac{1}{\sqrt{2}}\int{\frac{1}{\sqrt{x^2+\{\sqrt{\frac{3}{2}}\}^2}}}\,dx+\frac{1}{7}\int{\frac{1}{x^2+\{\sqrt{\frac{3}{7}}\}^2}}\,dx$


=$\frac{1}{\sqrt{2}}log(\sqrt{x^2+\{\sqrt{\frac{3}{2}}\}^2}+x)+\frac{1}{7}\sqrt{\frac{7}{3}}tan^{-1}\frac{x}{\sqrt{\frac{3}{7}}}+C$


= $tex]\frac{1}{\sqrt{2}}log(\sqrt{x^2+\{\sqrt{\frac{3}{2}}\}^2}+x)+\frac{1}{\sqrt{21}}tan^{-1}\frac{x}{\sqrt{\frac{3}{7}}}+C$