Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

 

dy/dx+yx^-1=tan^-1(x). Let p(x)=x^-1 and q(x)=tan^-1(x).

Multiply through by an unknown function u(x):

u(x)dy/dx+u(x)yp(x)=u(x)q(x).

d/dx(uy)=udy/dx+ydu/dx or u(x)dy/dx+ydu/dx.

Comparing this expansion of the differentiation of a product, we have du/dx=u(x)p(x)=u(x)/x.

So, du/u=dx/x; integrating: ln(u)=ln(x)+k, where k is a constant which can be written k=ln(a) and u(x)=ax, where a is a constant.

Now we have d/dx(axy)=axdy/dx+ay=a(xdy/dx+y)=ax(dy/dx+y/x)=axtan^-1(x).

Integrating wrt x axy=integral(axtan^-1(x)dx), xy=integral(xtan^-1(x)dx).

A slight diversion:

(1) integral(tan^-1(x)dx). Let r=tan^-1(x), x=tan(r), dx=sec^2(r)dr;

integral(tan^-1(x)dx)=integral(rsec^2(r)dr);

let U=r, dU=dr, let dV=sec^2(r)dr, V=tan(r);

d/dr(UV)=UdV/dr+VdU/dr=rsec^2(r)+tan(r); UV=rtan(r)=integral(rsec^2(r)dr)+integral(sin(r)/cos(r).dr).

Therefore, integral(rsec^2(r)dr)=rtan(r)-integral(sin(r)/cos(r).dr)=rtan(r)+ln(cos(r)).

tan(r)=x so cos(r)=1/sqrt(1+x^2). Substituting for r:

integral(tan^-1(x)dx)=xtan^-1(x)+ln(1/sqrt(1+x^2))=xtan^-1(x)-(1/2)ln(1+x^2).

(2) integral(ln(1+x^2)dx). Let U=ln(1+x^2), dU=2x/(1+x^2); let dV=dx, V=x.

d/dx(UV)=UdV/dx+VdU/dx; UV=xln(1+x^2)=integral(ln(1+x^2)dx)+integral(2x^2/(1+x^2).dx)=

integral(ln(1+x^2)dx)+integral(((2x^2+2)/(1+x^2)-2/(1+x^2))dx)=

integral(ln(1+x^2)dx)+integral(2dx)-integral(2/(1+x^2).dx)=

integral(ln(1+x^2)dx)+2x-2tan^-1(x).

Therefore, integral(ln(1+x^2)dx)=xln(1+x^2)-2x+2tan^-1(x).

Back to the problem:

integral(xtan^-1(x)dx): let u=x, du=dx; dv=tan^-1(x)dx, v=xtan^-1(x)-(1/2)ln(1+x^2).

d/dx(uv)=udv/dx+vdu/dx;

uv=x(xtan^-1(x)-(1/2)ln(1+x^2))=

integral(xtan^-1(x)dx)+integral((xtan^-1(x)-(1/2)ln(1+x^2))dx)=

2*integral(xtan^-1(x)dx)-(1/2)integral(ln(1+x^2)dx)=

2*integral(xtan^-1(x)dx)-(1/2)(xln(1+x^2)-2x+2tan^-1(x)).

Therefore:

2*integral(xtan^-1(x)dx)=x(xtan^-1(x)-(1/2)ln(1+x^2))+(1/2)(xln(1+x^2)-2x+2tan^-1(x)).

2*integral(xtan^-1(x)dx)=x(xtan^-1(x)-(1/2)ln(1+x^2))+(1/2)(xln(1+x^2)-2x+2tan^-1(x))=

x^2tan^-1(x)-x+tan^-1(x)=(tan^-1(x))(1+x^2)-x.

Therefore, integral(xtan^-1(x)dx)=(1/2)((1+x^2)tan^-1(x)-x) and

xy=(1/2)((1+x^2)tan^-1(x)-x) and y=(1/2x)((1+x^2)tan^-1(x)-x).
 

by Top Rated User (1.2m points)

Related questions

13 answers
asked Jan 11, 2014 in Algebra 2 Answers by Jennifer A. Cascaño Level 12 User (101k points) | 3.6k views
1 answer
1 answer
asked Jul 19, 2015 in Calculus Answers by anonymous | 585 views
1 answer
asked Feb 14, 2014 in Calculus Answers by Jennifer A. Cascaño Level 12 User (101k points) | 1.3k views
1 answer
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,516 questions
100,279 answers
2,420 comments
732,376 users