## anu143 3 years ago wts d integration of (x^2+a^2)dy/dx=(y+b)(x+sqrt(x^2+a^2))

1. him1618

you need to solve the diff eqn?

2. anu143

ya by variable separable

3. him1618

x wali sari expressions ek side karlo

4. anu143

its nt workin :(

5. him1618

you have dy / (y+b) = (x+sqrt(x^2 + a^2))dx / (x^2 + a^2) isnt it?

6. anu143

correct

7. him1618

so now wheres the problem?

8. anu143

m nt gettin hw 2 go ahead

9. him1618

ok the left hand side is clear to you isnt it??

10. anu143

yup d right one s mes

11. him1618

so fr the right side break the fraction into 2 parts

12. him1618

$\frac{x}{x^2 + a^2} + \frac{1}{\sqrt(x^2 + a^2)}$

13. him1618

ab aaya?

14. anu143

i gt it thanks:)

15. anu143

u on fb??

16. anu143

can i hav ur id??

17. hartnn

O.o

18. UnkleRhaukus

$(x^2+a^2)\frac{\text dy}{\text dx}=(y+b)(x+\sqrt{x^2+a^2})$ $\frac{\text dy}{y+b}=\frac{x+\sqrt{x^2+a^2}}{x^2+a^2}{\text dx}$ $\int\frac{\text dy}{y+b}=\int\frac{x+\sqrt{x^2+a^2}}{x^2+a^2}{\text dx}$

19. UnkleRhaukus

then uses these standard integrals for $$\alpha,\beta,\gamma\in\mathbb R$$ $\int\frac{\text du}{u+\alpha}=\ln|u+\alpha|$ $\int\frac{v}{v^2+\beta^2}\text dv=\frac12\ln|v^2+\beta^2|$ $\int\frac1{\sqrt{w^2\pm \gamma^2}}\text dw=2\sqrt{ x\pm\gamma}=$