Solve: int (4t^2 + 10 / (t^2 + 3)(t^2+4)) * dt\(\int\frac{4t^2+10}

1.	Solve: int (4t^2 + 10 / (t^2 + 3)(t^2+4)) * dt\(\int\frac{4t^2+10}{(t^2+3)(t^2+4)}dt\)
Answer» \(\int\frac{4t^2+10}{(t^2+3)(t^2+4)}dt\) = \(\int\frac{4(t^2+3)+2((t^2+4)-(t^2+3))}{(t^2+3)(t^2+4)}dt\) \(=\int(\frac4{t^2+4}-\frac2{t^2+3}+\frac3{t^2+4})dt\) \(=\int\frac{6}{t^2+4}dt-\int\frac{2}{t^2+3}dt\) \(=\frac62tan^{-1}\frac{t}2-\frac2{\sqrt3}tan^{-1}\frac{t}{\sqrt3}+c\) = 3 tan^-1\(\frac{t}2\) \(-\frac{2}{\sqrt3}tan^{-1}\frac{t}{\sqrt3}+c\)

Solve: int (4t^2 + 10 / (t^2 + 3)(t^2+4)) * dt\(\int\frac{4t^2+10}{(t^2+3)(t^2+4)}dt\)

Answer»

\(\int\frac{4t^2+10}{(t^2+3)(t^2+4)}dt\) = \(\int\frac{4(t^2+3)+2((t^2+4)-(t^2+3))}{(t^2+3)(t^2+4)}dt\)

\(=\int(\frac4{t^2+4}-\frac2{t^2+3}+\frac3{t^2+4})dt\)

\(=\int\frac{6}{t^2+4}dt-\int\frac{2}{t^2+3}dt\)

\(=\frac62tan^{-1}\frac{t}2-\frac2{\sqrt3}tan^{-1}\frac{t}{\sqrt3}+c\)

= 3 tan^-1\(\frac{t}2\) \(-\frac{2}{\sqrt3}tan^{-1}\frac{t}{\sqrt3}+c\)