Question

15. (r2 + 1) log x

Answer 1

The integral is given as follows,

I= ∫ ( x 2 +1 )logxdx

Rewrite the given integral as,

I= ∫ x 2 logx + ∫ logxdx I= I 1 + I 2

Assume, I 1 = ∫ x 2 logxdx

Use integration by parts rule. Consider logx as first function and x 2 as second function.

I 1 = ∫ x 2 logxdx I 1 =logx ∫ x 2 dx − ∫ ( d dx logx ∫ x 2 dx )dx I 1 = x 3 logx 3 − ∫ 1 x × x 3 3 dx

On further integrating, we get,

I 1 = x 3 logx 3 − 1 3 ∫ x 2 dx I 1 = x 3 logx 3 − 1 9 x 3 + C 1

Assume, I 2 = ∫ logxdx

Consider logxas first function and 1as second function. Use integration by parts.

I 2 = ∫ logx×1dx I 2 =log ∫ dx − ∫ ( d dx logx ∫ dx )dx

On further integrating, we get,

I 2 =xlogx− ∫ 1 x ×xdx I 2 =xlogx− ∫ dx I 2 =xlogx−x+ C 2

Substitute the values of I 1 and I 2 in I,

I= I 1 + I 2 = x 3 logx 3 − 1 9 x 3 + C 1 +xlogx−x+ C 2 = x 3 logx 3 +xlogx− x 3 9 −x+C I=( x 3 3 +x )logx− x 3 9 −x+C

Thus, the integration of ∫ ( x 2 +1 )logxdx is ( x 3 3 +x )logx− x 3 9 −x+C.