Advertisements
Advertisements
Question
Evaluate ; `int x^2/(x^6 - 4x^3 + 13)` dx
Solution
Let I = `int x^2/(x^6 - 4x^3 + 13)` dx
= `int x^2/((x^3)^2 - 4x^3 + 13)` dx
Put x3 = t
3x2dx = dt
x2dx = `dt/3`
I = `int ((dt/3))/(t^2 - 4t + 13)`
= `1/3 int ((dt))/[(t^2 - 4t + 4) - 4 + 13]`
= `1/3 1/[(t - 2)^2 + 3^2) dt`
= `1/3 . 1/3 tan^-1 ((t -2)/3) + c`
`int 1/(x^2 + a^2) dx = 1/a tan ^-1 (x/a) + c`
I = `1/9 tan^-1 ((x^3 - 2)/3) + c`
APPEARS IN
RELATED QUESTIONS
evaluate : `int xcosxdx`
Evaluate :`int_0^(pi/2)dx/(1+cotx)`
Evaluate : `int (2"x" + 1)/(("x" + 1)("x"-2)) "dx"`
If the function f is continuous at x = 2 and x = 4 then find the values of a and b.
Where f(x) = x2 + ax + b, x < 2
= 3x + 2, 2 ≤ x ≤ 4
= 2ax + 5b, 4 < x
`int (2x + 1)/((x + 1) ( x - 2))` dx
Evaluate : `int x cosx dx`
Evaluate : `∫((1 + logx))/(x(2 + logx)(3 + logx)`dx
