Question

hlo..step by step (integration)​

Answer 1

Hey mate, here is your AnsweR,

Question :

• $\: \sf \int \frac{3x + 1}{ \sqrt{5 - 2x - x {}^{2} } } dx$

Step by step Explanation :

$\bf \underbrace{ \pmb{Problem \: : }} \\\\\ \bull \: \: \: \sf \int \frac{3x + 1}{ \sqrt{5 - 2x - x {}^{2} } } dx$

$➠ \: \: \tt \int \bigg( - \frac{3( - 2x - 2)}{2 \sqrt{ - x {}^{2} - 2x + 5 } } - \frac{2}{ \sqrt{ - x {}^{2} - 2x + 5 } } \bigg) \text{dx} \\\\\ ➠ \: \: \tt3 \int \frac{x + 1}{ \sqrt{ - x {}^{2} - 2x + 5 } } dx - 2 \int \frac{1}{ \sqrt{ - x {}^{2} - 2x + 5} } \text dx$

$\bf \underbrace{ \pmb{Now \:Solving \: : }} \\\\\ \bull \: \: \: \sf \int \frac{x + 1}{ \sqrt{ - x {}^{2} - 2x + 5} } dx$

$\tt➥ \: \: - \frac{1}{2} \int \frac{1}{ \sqrt{u} } \text du \\\\\ \tt➥ \: \:2 \sqrt{u}$

$\bf \underbrace{ \pmb{ Plug \: in \: solved \: integrals\: : }} \\\\\ \bull \: \: \: \sf - \frac{1}{2} \int \frac{1}{ \sqrt{u} }du$

$\tt➲ \: \: - \sqrt{ - x { }^{2} - 2x + 5 }$

$\bf \underbrace{ \pmb{Now \:Solving \: : }} \\\\\ \bull \: \: \: \sf \int \frac{ 1}{ \sqrt{ - x {}^{2} - 2x + 5} } dx$

$➯ \: \int \frac{1}{ \sqrt{6 - (x + 1) {}^{2} } } \text {dx} \\\\\ \: \tt➯ \: \int \frac{ \sqrt{6} }{ \sqrt{6 - 6u{}^{2} } } \text {dx} \\\\\ \: \tt➯\int \frac{1}{ \sqrt{1 - u {}^{2} } } \\\\\ \: \tt➯ \text { \: arcsin(u)}\\\\\ \tt➯ { \text { \: arcsin} \bigg( \frac{x + 1}{ \sqrt{6} } \bigg)}$

$\bf \underbrace{ \pmb{ Plug \: in \: solved \: integrals\: : }} \\\\\ \bull \: \: \: \sf3 \int \frac{x + 1}{ - x {}^{2} - 2x + 5} \text{dx} - 2 \int \frac{1}{ \sqrt{ - x {}^{2} - 2x + 5} } \text{dx}$

$✰\:\: \underline{ \underline{ \boxed {\frak{➣ \: \:{ - 2 \text{arcsin} \bigg( \frac{x + 1}{ \sqrt{6} } \bigg) - 3 \sqrt{ - x {}^{2} - 2x + 5 } + \bf{C}}}}}}\:\:✰$

Please mark my answer as brainliest answer.