Question

Evaluate the following integrals:
(i) ∫ x ln(x + 1)dx

(ii) ∫ ((sin³ (In x) cos²(In x)) / (x)) dx

Answer 1

i. Let $I=\int xln[1+x]dx$

Using integration by parts : $I=ln[1+x]\int xdx - \int[(d[ln[1+x]/dx) * \int xdx]dx$

$I=ln[1+x]*(x^2/2) -\int[(1/(1+x)*1*(x^2/2)]dx$

or, $I=(x^2/2)*ln[1+x] - (1/2)*\int$ $[x^2/(1+x)]dx$

Let us calculate $\int [x^2/(1+x)]dx$

$=\int(x^2-1+1)/(1+x)dx$

$=\int[(x-1)(x+1)/(x+1)+(1/(1+x))]dx$

$=\int[(x-1) +(1/(1+x))]dx$

$=\int(x-1)dx+\int1/$ $(1+x)dx$

$Let 1+x$ =$z$

therefore, $dx=dz$

therefore , $I=\int(x-1)dx+\int(1/z)dz$

$=(x^2/2)-x+lnz$

Putting $z=1+x$

$I=(x^2/2)-x+ln[1+x]$

Putting this back to $I$ we get:

$I=(x^2/2)ln[1+x] -(1/2)*[x^2/2-x+ln[1+x]]+C$

$=(x^2/2)ln[1+x]-(x^2/4)+(x/2)-(1/2)*ln[1+x]+C$

where C is an integration constant (Answers)

ii. Let $I=\int[sin^3(lnx)cos^2(lnx)]/xdx$

Let $lnx=z$

therefore, $(1/x)dx=dz$

therefore, $I=\int[sin^3zcos^2z]dz$

$=\int[sin^2zcos^2sinz]dz$

Let $cosz=u$

$-sinzdz=du$

therefore, $I=\int-(1-u^2)*u^2du$

$=\int(-u^2+u^4)du$

$=-u^3/3+u^5/5 +c$ where C is an integration constant

Putting back $u=cosz$ we get

$I=-cos^3z/3+cos^5z/5 + C$

Again putting back $z=lnx$ we get:

$I=-cos^3(lnx)/3 + cos^5(lnx)/5 +C (Answer)$