also:

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.

To integrate the function ∫tan^(-1)(y) dy, we can use the method of integration by parts.

First, we can rewrite the integral as ∫1 * tan^(-1)(y) dy.

Next, we choose u = tan^(-1)(y) and dv = 1 * dy.

Taking the derivatives and integrals, we have du = (1/(1+y^2)) dy and v = y.

Using the integration by parts formula ∫u dv = uv - ∫v du, we can substitute our values:

∫tan^(-1)(y) dy = uv - ∫v du
= y * tan^(-1)(y) - ∫y * (1/(1+y^2)) dy

Simplifying the integral part, we have:

∫y/(1+y^2) dy

Integration of this type of function can be done by applying the substitution method or directly integrating it using trigonometric identities.

Assuming we directly integrate it, we can rewrite the integral as:

∫y/(1+y^2) dy = 1/2 ∫(2y)/(1+y^2) dy

Using the substitution method, let u = 1+y^2 and du = 2y dy. Solving for y dy, we have dy = du/(2y).

Substituting these values, we get:

1/2 ∫(2y)/(1+y^2) dy = 1/2 ∫(1/u) du

Now, integrating ∫(1/u) du yields ln|u| + C, where C is the constant of integration.

Therefore, the integral becomes:

∫tan^(-1)(y) dy = y * tan^(-1)(y) - 1/2 ln|1+y^2| + C

And that is the final result of the integration.

To evaluate the integral of arctan(y) dy using integration by parts, you can follow these steps:

1. Start by writing arctan(y) dy as a product of two functions: arctan(y) and dy.

2. Apply the product rule of differentiation, which states that the derivative of the product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function.

3. Use the product rule to differentiate the product arctan(y) dy. The derivative of arctan(y) is 1/(1+y^2), and the derivative of dy is 0.

4. Now you have arctan(y) dy = d[y arctan(y)] - y d[arctan(y)], as the derivative of dy is zero.

5. Simplify the expression by substituting the derivative of arctan(y) with 1/(1+y^2).

6. The integral of arctan(y) dy becomes the integral of d[y arctan(y)] - y d[arctan(y)].

7. Integrate both terms separately. The integral of d[y arctan(y)] is y arctan(y) + C, where C is the constant of integration.

8. The integral of -y d[arctan(y)] can be evaluated by substitution or integration by parts again. Let's use integration by parts again.

9. In the new integral -y d[arctan(y)], let u = arctan(y) and dv = -y dy. Then, du = 1/(1+y^2) dy and v = -1/2 y^2.

10. Apply integration by parts to the integral -y d[arctan(y)]. The formula for integration by parts is ∫ u dv = uv - ∫ v du.

11. Using the formula, the integral -y d[arctan(y)] becomes (-1/2)y^2 arctan(y) + (-1/2)∫ (1/(1+y^2)) y^2 dy.

12. The integral ∫ (1/(1+y^2)) y^2 dy can be simplified using a trigonometric substitution or by recognizing it as the integral of a rational function. The result is 1/2 ln(1+y^2) + D, where D is another constant of integration.

13. Putting it all together, the integral of arctan(y) dy is y arctan(y) + (-1/2)y^2 arctan(y) - 1/2 ln(1+y^2) + C.

So, the final result is y arctan(y) - (1/2)y^2 arctan(y) - 1/2 ln(1+y^2) + C, where C is the constant of integration.