In this case, since the substitution is given in the question, we just need to carry out the procedure for substitution.
Since `u=a+bx` , then `du=bdx` so `dx=1/bdu` and `x=1/b(u-a)` . This means the integral becomes:
`int x/(a+bx)^{3/2}dx`
`=1/b^2int {u-a}/u^{3/2}du` now split the integral into two integrals
`=1/b^2int u/u^{3/2}du-a/b^2int 1/u^{3/2}du` simplify...
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In this case, since the substitution is given in the question, we just need to carry out the procedure for substitution.
Since `u=a+bx` , then `du=bdx` so `dx=1/bdu` and `x=1/b(u-a)` . This means the integral becomes:
`int x/(a+bx)^{3/2}dx`
`=1/b^2int {u-a}/u^{3/2}du` now split the integral into two integrals
`=1/b^2int u/u^{3/2}du-a/b^2int 1/u^{3/2}du` simplify the integrands
`=1/b^2intu^{-1/2}du-a/b^2int u^{-3/2}du` now use power law
`=2/b^2u^{1/2}+{2a}/b^2u^{-1/2}+C` where C is the constant of integration
`={2(u+a)}/{b^2u^{1/2}}+C`
The integral evaluates to `{2(u+a)}/{b^2u^{1/2}}+C` .