# Solve Calc 2 Integral 0 to h: 1/((h-r)^2+r^2))

• t8ened
In summary, the poster is struggling with solving the integral 1/((h-r)^2+r^2))dr from 0 to h. They suggest using partial fractions or a trig sub, and provide some steps to simplify the expression. They also give a hint to complete the square and find the antiderivative.
t8ened

## Homework Statement

this integral owned me, can anyone tell me what to do to solve it/ how to solve it or a step by step solution. from 0 to h, 1/((h-r)^2+r^2))dr

## The Attempt at a Solution

Im thinking partial factions or trig sub...

Let's look at the denominator.
(h - r)^2 + r^2 = h^2 - 2hr + r^2 + r^2
= 2r^2 - 2hr + h^2
= 2(r^2 - rh) + h^2

Now, complete the square in the first expression to get
2(r - A)^2 + B^2 (you'll have to figure out A and B)

The antiderivative of the expression above is K * arctan(something) + C.

Enough of a hint?

## What is the integral of 1/((h-r)^2+r^2) from 0 to h?

The integral of 1/((h-r)^2+r^2) from 0 to h is arctan((h-r)/r) evaluated at h and 0, which simplifies to arctan(h/r) - arctan(-r/r) or arctan(h/r) + pi/2.

## How do you solve the integral of 1/((h-r)^2+r^2) from 0 to h?

To solve this integral, you can use trigonometric substitution. Let r = h*tan(u) and solve for dr in terms of u. Then substitute these expressions into the integral and simplify to get the final answer.

## What is the domain of the function 1/((h-r)^2+r^2)?

The domain of the function 1/((h-r)^2+r^2) is all real numbers except for when the denominator is equal to 0. In other words, the domain is all real numbers except for r = h.

## What is the range of the function 1/((h-r)^2+r^2)?

The range of the function 1/((h-r)^2+r^2) is all real numbers greater than or equal to 0. This is because the numerator is always 1, and the denominator is always positive, so the function can never be negative.

## How does the graph of 1/((h-r)^2+r^2) from 0 to h look like?

The graph of 1/((h-r)^2+r^2) from 0 to h is a smooth, continuous curve. It starts at 0 when r = 0, increases as r increases, and approaches infinity as r approaches h. There are no vertical asymptotes, and the curve is always above the x-axis.

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