Question

integrate
1)√x-1/√x
2)√(3x-4)³
3)(x+1/x)³​

Answer 1

I=∫x(1−x)3dx ; Substitute 1−x=t ; dx=−dt

=−∫(1−t)t3dt

=−∫(t3−t4)dt

=−(t44−t55)+C

=(1−x)55−(1−x)44+C

The method used above is too long.Simpler is,

I=∫x(1−x)3dx

=∫{1−(1−x)}(1−x)3dx

=∫{(1−x)3−(1−x)4}dx

=−(1−x)44+(1−x)55+C

=(1−x)55−(1−x)44+C