Monday, August 17, 2015

`int_1^2 (ln(x))^2/x^3 dx` Evaluate the integral

`int_1^2(ln(x))^2/x^3dx`


If f(x) and g(x) are differentiable function, then


`intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx`


If we rewrite f(x)=u and g'(x)=v, them


`intuvdx=uintvdx-int(u'intvdx)dx`


Using the above method of integration by parts,


Let's first evaluate the indefinite integral,


`int(ln(x))^2/x^3dx=(ln(x))^2int1/x^3dx-int(d/dx(ln(x))^2int(1/x^3)dx)dx`


`=(ln(x))^2(x^(-3+1)/(-3+1))-int(2ln(x)(1/x)(x^(-3+1)/(-3+1)))dx`


`=(ln(x))^2(x^(-2)/-2)-int(2ln(x)(1/x)(x^(-2))/-2)dx`


`=-1/(2x^2)(ln(x))^2+int(ln(x)/x^3)dx`


again applying integration by parts


`=-1/(2x^2)(ln(x))^2+ln(x)*int(1/x^3)dx-int(d/dx(ln(x)int(1/x^3)dx)dx`


`=-1/(2x^2)(ln(x))^2+ln(x)*(x^(-3+1)/(-3+1))-int(1/x)((x^-3+1)/(-3+1))dx`


`=-1/(2x^2)(ln(x))^2+ln(x)*(x^(-2)/-2)-int(1/(-2x^3))dx`


`=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)+1/2int(1/x^3)dx`


`=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)+1/2(x^(-3+1)/(-3+1))`


`=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)-1/(4x^2)`


add a constant C to the solution,


`=-1/(2x^2)((ln(x))^2+ln(x)+1/2)+C`


Now let's evaluate definite integral,


`int_1^2(ln(x))^2/x^3dx=[-1/(2x^2)((ln(x))^2+ln(x)+1/2)]_1^2`


`=[-1/(2*2^2)((ln(2))^2+ln(2)+1/2)]-[-1/(2*1^2)((ln(1))^2+ln(1)+1/2)]`


`=[-1/8(ln(2))^2-ln(2)/8-1/16+1/4]`


`=3/16-(ln(2))^2/8-ln(2)/8`


`int_1^2(ln(x))^2/x^3dx`


If f(x) and g(x) are differentiable function, then


`intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx`


If we rewrite f(x)=u and g'(x)=v, them


`intuvdx=uintvdx-int(u'intvdx)dx`


Using the above method of integration by parts,


Let's first evaluate the indefinite integral,


`int(ln(x))^2/x^3dx=(ln(x))^2int1/x^3dx-int(d/dx(ln(x))^2int(1/x^3)dx)dx`


`=(ln(x))^2(x^(-3+1)/(-3+1))-int(2ln(x)(1/x)(x^(-3+1)/(-3+1)))dx`


`=(ln(x))^2(x^(-2)/-2)-int(2ln(x)(1/x)(x^(-2))/-2)dx`


`=-1/(2x^2)(ln(x))^2+int(ln(x)/x^3)dx`


again applying integration by parts


`=-1/(2x^2)(ln(x))^2+ln(x)*int(1/x^3)dx-int(d/dx(ln(x)int(1/x^3)dx)dx`


`=-1/(2x^2)(ln(x))^2+ln(x)*(x^(-3+1)/(-3+1))-int(1/x)((x^-3+1)/(-3+1))dx`


`=-1/(2x^2)(ln(x))^2+ln(x)*(x^(-2)/-2)-int(1/(-2x^3))dx`


`=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)+1/2int(1/x^3)dx`


`=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)+1/2(x^(-3+1)/(-3+1))`


`=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)-1/(4x^2)`


add a constant C to the solution,


`=-1/(2x^2)((ln(x))^2+ln(x)+1/2)+C`


Now let's evaluate definite integral,


`int_1^2(ln(x))^2/x^3dx=[-1/(2x^2)((ln(x))^2+ln(x)+1/2)]_1^2`


`=[-1/(2*2^2)((ln(2))^2+ln(2)+1/2)]-[-1/(2*1^2)((ln(1))^2+ln(1)+1/2)]`


`=[-1/8(ln(2))^2-ln(2)/8-1/16+1/4]`


`=3/16-(ln(2))^2/8-ln(2)/8`


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