∫(上限1下限0) x/√(4+5x)要怎麼算~~ 明天要考的內容,解不開?

Sol
A=∫(0 to 1)_x/√(4+5x)dx
Set 4+5x=y
5dx=dy
25A=∫(0 to 1)_25x/√(4+5x)dx
=∫(0 to 1)_5x/√(4+5x)d(5x)
=∫(4 to 9)_(y－4)/y^(1/2)dy
=∫(4 to 9)_y^(1/2)－4y^(－1/2)dy
=y^(3/2)/(3/2)－4y^(1/2)/(1/2)｜(4 to 9)
=[(2/3)*9^(3/2)－8*9^(1/2)]－[(2/3)*4^(3/2)－8*4^(1/2)]-
=[(2/3)*27－8*3]－[(2/3)*8－8*2]-
=(18－24)－(16/3－16)
-=18－24－16/3+16
=14/3
A=14/75