# Solve for y: APC.7.3.01 Int Subs

• MHB
• karush
In summary, the conversation discusses solving for $y$ in the equation $u=1+\ln(y^2)$ and finding the integral of $\frac{1}{y[1+\ln(y^2)]}$. The steps involved include using substitution, algebra, and the properties of logarithms and integrals. The final answer is $y=e^{\frac{u-1}{2}}$ and the output should be preceded by "In summary, ".
karush
Gold Member
MHB

ok hope the image is readable
kinda odd with the y is in the denominator
but was going to start with $u=1+\ln y^2 \quad y=??$
not sure ?

no need to solve for $y$ ...

$u = 1+ \ln(y^2) = 1+2\ln{y} \implies du = \dfrac{2}{y} \, dy$

$\displaystyle \dfrac{1}{2} \int \dfrac{du}{u} = \dfrac{1}{2}\ln|u|+ C$

$\displaystyle \dfrac{1}{2} \ln|1+\ln(y^2)| + C$

ok I don't think I see what happened between the first and second step

$$dy=\frac{y}{2}du$$

never mind I see what it is

Last edited:
$\int \dfrac{dy}{y[1+\ln(y^2)]} = \dfrac{1}{2} \int \dfrac{1}{{\color{blue}1+2\ln{y}}} \cdot {\color{red}\dfrac{2}{y} \, dy} = \dfrac{1}{2} \int \dfrac{1}{{\color{blue}u}} \, {\color{red}du}$

Yes, you don't need to solve for y. But if you did it is just algebra:
$u= 1+ ln(y^2)= 1+ 2ln(y)$
$u- 1= 2 ln(y)$
$\frac{u-1}{2}= ln(y)$
$e^{\frac{u-1}{2}}= y$.

## What does APC.7.3.01 Int Subs stand for?

APC.7.3.01 Int Subs stands for "Advanced Placement Calculus, Section 7.3.01: Integration by Substitution."

## What is the purpose of solving for y in this equation?

The purpose of solving for y in this equation is to find the value of y that satisfies the given equation and to express the equation in terms of the variable y.

## What are the steps to solve for y in this equation?

The steps to solve for y in this equation are as follows: 1) Identify the variable y in the equation. 2) Isolate y on one side of the equation by performing algebraic operations on both sides. 3) Simplify the equation to solve for y. 4) Check the solution by substituting the value of y back into the original equation.

## What are some common mistakes when solving for y in this equation?

Some common mistakes when solving for y in this equation include: 1) Forgetting to perform the same operation on both sides of the equation. 2) Making a mistake while simplifying the equation. 3) Forgetting to check the solution by substituting it back into the original equation. 4) Misidentifying the variable y in the equation.

## How can I check my solution for y in this equation?

You can check your solution for y in this equation by substituting the value of y back into the original equation and solving for both sides to see if they are equal. If they are equal, then your solution is correct. If they are not equal, then you may have made a mistake during the solving process.

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