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Script To Hack Instagram Accounts

And then the next steps start. The user is told to locate the picture he believes is the original image that was used to impersonate the owner. “After that you need to log into your Instagram account, then you can continue to the next steps,” the tutorial continues.

This is a very powerful tool for automatically searching inside your Instagram account. This is an open source tool. Open this, It will give you an error message as there is no Instagram account in this, we have to login our account and you can see that we have found our Instagram account there.
Now we are going to hack this account.
Click on Login and click on Instagram. A pop up will appear, enter your username and password. Then click on Ok. And you are done.
Then click on get and you will get the hacked account. Take a look at the screen, this is how you will hack any Instagram account.

If everything goes fine, a pop up with a message will appear like “Your password is changed”
So, you have to change it in your Instagram account.

Now, to enter the Instagram site, we need to click on “install”

Instructions
How to Hack Instagram accounts with InstaHacker
How to use InstaHacker script
Step 1: Install the Instagram Hack Script into Termux
Step 2: Connect Instagram Accounts on Termux
Step 3: Hack Instagram Accounts
Step 4: Delete fake accounts from instagram
Step 5: How to Hack Instagram accounts with InstaHacker
Step 6: Exit InstaHacker
Note: Make sure that you have installed Termux and Termux apk into your Android phone.Q:

Provide solution to this injunction $\int_{0}^{1} x\sqrt[3]{x}dx$

I’m doing a practice question and in it it asks to provide solutions to this integral:
$$\int_{0}^{1} x\sqrt[3]{x}dx$$
My approach:
The only possible substitution is $\sqrt[3]{x}\mapsto y$ thus $\int_{0}^{1} x^{\frac{2}{3}}dx=\int_{0}^{1} y^{2}dy=\frac{1}{3}\int_{0}^{1} y^{3}dy$.
But this is not possible to integrate by parts:
$$\frac{1}{3}\int_{0}^{1} y^{3}dy=\frac{1}{3}\frac{y^{4}}{4}|_{0}^{1}+\frac{1}{3}\int_{0}^{1} (3\cdot y^{2})dy=\frac{1}{3}\frac{y^{4}}{4}|_{0}^{1}-\frac{1}{3}\int_{0}^{1} y^{2}dy=\frac{1}{3}\frac{y^{4}}{4}|_{0}^{1}+\frac{2}{3}\frac{y^{3}}{3}|_{0}^{1}$$
This can’t be simplified because this further implies $\int_{0}^{1} y^{2}dy eq \frac{2}{3}y^{3}|_{0}^{1}$