Question

8x+13/√4x+7,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

HELLO DEAR,

GIVEN:-
∫(8x + 13)/√(4x + 7).dx

put 4x + 7 = t => 8x = 2t - 14
4.dx = dt
=> dx/4.dt

therefore, ∫(2t - 14 + 13)/√t.4dt

=> I = 1/4∫2t/√t.dt - 1/4∫1/√t.dt

=> I = 1/2∫√t.dt - 1/4∫t^{-1/2}.dt

=> I = 1/2*(2/3)t^{3/2} - 1/4*(2)t^{1/2} + C

=> I = 1/3t^{3/2} - 1/2√t + C.

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hello,

Solution:

∫$\frac{8x+13}{\sqrt{4x+7} } dx$

let 4x+7 = t

so, 4 dx = dt

dx = dt/4

substitute these values, we get

= 1/4 ∫ $\frac{2t-1}{\sqrt{t} } dt$

apply linearity

= 1/4 ∫ $( 2\sqrt{t} -\frac{1}{\sqrt{t} }) dt$

= 1/ 2 ∫ √t dt -1/4 ∫ 1/√t  dt

apply power rule and do integration,

= $\frac{1}{2} \frac{2}{3} t^{\frac{3}{2} } -\frac{1}{4} 2\sqrt{t} +C\\ \\ = \frac{1}{3} t^{\frac{3}{2} } - \frac{1}{2} \sqrt{t}+C$

undo substitution, and simplify


∫$\frac{8x+13}{\sqrt{4x+7} } dx$ = $\frac{\sqrt{(4x+7)}(8x+11) }{6} +C$

Hope it helps you.