Calculus Problems?

Question

How to Integrate
1)x^2 tan^-1 x dx?(*x is not the power of tan)
2)xln(9+x^2)dx?

Answer 1

1)

∫x^2 tan^-(x) dx

integrating by parts

u = tan^-1(x) : du = 1/(1+x^2)

dv = x^2 : v = x^3/3

∫x^2 tan^-(x) dx = uv - ∫vdu

=>tan^-1(x)(x^3/3) -(1/3)∫x^3/(1+x^2) dx

write x^3/1+x^2 = x - x/(1+x^2) (dividing x^3 by (1+x^2)

=>tan^-1(x)(x^3/3) -(1/3)∫x - (x/(1+x^2) dx

=>(x^3/3)tan^-1(x) -(1/3)∫x dx + (1/3)∫1/2 (2x)/(1+x^2) dx

=>(x^3/3)tan^-1(x) -(1/3)∫x dx + (1/6)∫ (2x)/(1+x^2) dx

=>(x^3/3)tan^-1(x) -(1/3)x^2/2+ (1/6)ln(1+x^2)+c

=>(x^3/3)tan^-1(x) -(1/6)x^2 + (1/6)ln(1+x^2)+c

2)

∫x ln(9+x^2) dx

integrating by parts

u = ln(9+x^2) : du = (1/(9+x^2))*(2x)

dv = x : v = x^2/2

∫x ln(9+x^2) dx = uv - ∫v du

=>ln(9+x^2)(x^2/2) - ∫x^2(2x)/2(9+x^2) dx

=> ln(9+x^2)(x^2/2) - ∫x^3/(9+x^2) dx

=>ln(9+x^2)(x^2/2) - ∫x - (9x/(9+x^2)) dx

=>ln(9+x^2)(x^2/2) - ∫xdx + 9/2 ∫(2x/(9+x^2)) dx

=>ln(9+x^2)(x^2/2) - x^2/2 + 9/2 ln(9+x^2) + c

=>(x^2/2)ln(9+x^2) - x^2/2 + (9/2) ln(9+x^2) + c

Answer 2

One problem per question please.

∫[xln(9 + x²)] dx
Let
9 + x² = u
2x dx = du
x dx = du/2

(1/2)∫ln(u) du = (1/2)[u(ln u) - u) + C

= (1/2)[(9 + x²)ln(9 + x²) - (9 + x²)] + C

= (1/2)(9 + x²)[ln(9 + x²) - 1] + C

Answer 3

Use product rule in both problems.
x^2(1/(1+x^2) +2x(tan^-1(x))
= x^2/(1+x^2) +2x(tan^-1(x))

ln(9+x^2) + x *2x/(1+x^2)
= ln(9+x^2) + 2x^2/(1+x^2)