How do you evaluate the definite integral #int 8/(3+4x)# from #[0,1]#?

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Answer 1 · 2016-09-15T14:19:31

# int_0^1 8/(3+4x)dx=2ln(7/3), or, =ln(49/9).#

We know that, #int1/tdt=ln|t|+C...........(1)#

If we let, #t=3+4x, "then", dt=4dx.#

By #(1), "then", int1/(3+4x)*4dx=ln|3+4x|+C,# or,

#int 8/(3+4x)dx=2ln|3+4x|+C.#

Therefore,

#int_0^1 8/(3+4x)dx=2[ln|3+4x|]_0^1#

#=2[ln7-ln3]=2ln(7/3)#

#:. int_0^1 8/(3+4x)dx=2ln(7/3), or, =ln(49/9).#

Enjoy Mats.!