Saturday

December 20, 2014

December 20, 2014

Posted by **marsha** on Wednesday, May 23, 2007 at 5:34pm.

integral of tan^(-1)y dy

how is integration of parts used in that?

You write:

arctan(y)dy = d[y arctan(y)] -

y d[arctan(y)]

Here we again have used the product rule:

d(fg) = f dg + g df

You then use that:

d[arctan(y)] = 1/(1+y^2) dy

So, the integral becomes:

y arctan(y) - Integral of y/(1+y^2) dy

And:

Integral of y/(1+y^2) dy =

1/2 Log[1+y^2] + const.

**Answer this Question**

**Related Questions**

calculus - h(x)= integral from (1, 1/x) arctan(2t)dt part 1: let U= 1/x and du...

calculus - h(x)= integral from (1, 1/x) arctan(2t)dt part 1: let U= 1/x and du...

precal - The values of x that are solutions to the equation cos^(2)x=sin2x in ...

calculus - Let f be a function defined by f(x)= arctan x/2 + arctan x. the value...

check my arctan please - arctan(1/12)= 4.763 rounded to the nearest tenth degree...

Calculus - Note that pi lim arctan(x ) = ---- x -> +oo 2 Now evaluate / pi \ ...

Calc - Evaluate the integral (3x+4)/[(x^2+4)(3-x)]dx a. (1/2)ln(x^2+4) + ln|3-x...

limiting position of the particle - A particle moves along the x axis so that ...

Math - Arrange these in order from least to greatest: arctan(-sqrt3), arctan 0, ...

PRE-CALC - sin(arctan(-4/3) okay so i made arctan(-4/3) = x so im solving for ...