Question

Find the value of ∫∫ ^x⁄x^2 + y^2 dxdy.

(a) [ytan^(-1) (y)- ^1⁄2 ln⁡(1+y^2)]

(b) x [ytan^(-1) (y)- ^1⁄2 ln⁡(1+y^2)]

(c) y [xtan^(-1) (x)- ^1⁄2 ln⁡(1+x^2)]

(d) x [ytan^(-1) (y)- ^1⁄2 ln⁡(1+y^2)]

This question was posed to me in a job interview.

The doubt is from Double Integrals topic in section Multiple Integrals of Engineering Mathematics

Answer 1

Correct option is (d) x [ytan^(-1) (y)- ^1⁄2 ln⁡(1+y^2)]

Explanation: Add constant automatically

Given, \(\int\int \frac{x}{x^2+y^2} \,dxdy\)

\(\int x \int \frac{1}{x^2+y^2} \,dydx=\int x \frac{1}{x} tan^{-1}⁡(\frac{y}{x}) \,dy=\int tan^{-1}⁡(\frac{y}{x}) \,dy\)

\(\int tan^{-1}⁡(\frac{y}{x})\,dy = x\int tan^{-1}⁡(t)\,dt\)

Putting, x = tan(z),

We get, dz = sec^2⁡(z)dz,

x∫ zsec^2 (z)dz

By integration by parts,

x ∫ zsec^2 (z)dz=x[ztan(z)-log⁡(sec⁡(z))]= x[ytan^(-1) (y)- ^1⁄2 ln⁡(1+y^2)].