Question

Apply second substitution theorem evaluate

Integral 1 to 4 dt/(|t+4|√t)

Answer 1

$I=\int_1^4\frac {dt}{|t+4|\sqrt t}$

Substitution:

$u=\sqrt t$ ; $du=\frac12\frac{dt}{\sqrt t}$;

$u_1=\sqrt 1=1$ ; $u_2=\sqrt 4=2$ ;

$I=\int_1^4\frac {dt}{|t+4|\sqrt t}=2\int_1^4\frac {dt}{2|t+4|\sqrt t}=2\int_1^2\frac {du}{|u^2+4|}=$$2\int_1^2\frac {du}{u^2+4}$

Second substitution:

$v=\frac{u}{2}$; $dv=\frac{du}{2}$;

$v_1=\frac12$; $v_2=1$;

$I=4\int_{\frac12}^1\frac{dv}{4v^2+4}=\int_{\frac12}^1\frac{dv}{v^2+1}=\arctan{v}|_\frac12^1=$$\arctan{1}-\arctan{\frac12}=\frac{\pi}{4}-\arctan{\frac12} \approx0.32$