Question

Evaluate `int_1^2 (3x)/((9x^2 - 1))  "d"x`

Answer 1

Let I = `int_1^2 (3x)/((9x^2 - 1))  "d"x`

= `3int_1^2 x/(9x^2 - 1)  "d"x`

Put 9x² – 1 = t

∴ 18x dx = dt

∴ x dx = `1/18` dt

When x = 1, t = 9(1)² – 1 = 8

When x = 2, t = 9(2)² – 1 = 35

∴ I = `3int_8^35 1/"t"*"dt"/18`

= `1/6 int_8^35  "dt"/"t"`

= `1/6[log|"t"|]_8^35`

= `1/6(log 35 - log 8)`

∴ I = `1/6 log (35/8)`