Matematické Fórum

stuart clark · 27. 04. 2019 16:14

Finding $\displaystyle \int\frac{(x^2-1)\sin x+x\cos x}{(x^2+1)-2x\sin (2x)+(x^2-1)\cos(2x)}dx$

krakonoš

↑ stuart clark:
Hi Stuart Clark
$\int_{}^{}\frac{(x^{2}-1)sinx+x cosx}{(x^{2}+1)-2xsin(2x)+(x^{2}-1)cos(2x)}dx=$
$=\int_{}^{}\frac{x^{2}sinx-sinx+xcosx}{2\cdot (xcosx-sinx)^{2}}dx=$
$=\frac{1}{2}\cdot \frac{x}{xcosx-sinx}+C$

stuart clark · 29. 04. 2019 15:19

Thanks ↑ krakonoš:

Matematické Fórum

#1 27. 04. 2019 16:14

Indefinite Integration

#2 27. 04. 2019 20:27 — Editoval krakonoš (27. 04. 2019 20:50)

Re: Indefinite Integration

#3 29. 04. 2019 15:19

Re: Indefinite Integration

Zápatí