# What is the intergral of 6.x^2 + 2/x^2 + 5 with respect to x?

**What is the intergral of 6.x^2 + 2/x^2 + 5 with respect to x?**

To answer a question like this you would **intergrate each phrase** of the addition in turn with** respect to x**.

To intergrate a single *c.x^n* phrase with respect to *x*;

-divide *c* by *(n + 1)*

-raise *n* by *1*

Giving you *(c/(n+1)).x^(n+1).*

**For example the first phrase:** *6x^2;*

*c* is *6*, *n* is *2*;

so the intergral of *6.x^**2* with respect to *x* is* (6/(2 + 1)).x^(2+1) = 2.x^3.*

**For the second phrase,** *2x^-2* which is equivalent to *2/x^2*,

*c* is *2*, and n is *-2*;

so the intergral of the phrase with respect to *x* is *(2/(-2+1)).x^(-2+1) = -2.x^-1. *

**For the final phrase,** *5*, which is in the form of simply '*c*', intergrates easily;

*c* is *5*, n is *0*;

So the intergral of this phrase with respect to *x* is *(5/(0+1)).x^(0+1) = 5.x^1 = 5.x.*

This can be done quickly by thinking of the intergral of a constant *c* as simply *c.x.*

The full intergral of the entire question *6.x^2 + 2.x^-2 + 5* with respect to *x* is:

*2.x^3 -2.x^-1 + 5.x + c*

You may notice the extra *c* added on the end. The *c* occurs because the differentation of any constant *c* is *0*, so there may be any value of *c* in the intergral. This c is called the arbitrary **constant of intergration**, whose value can be found with additional information, producing a **particular intergral**. The important thing to remember though is that when you intergrate a function, **include + c on the end!**

Note: I can't seem to get superscript to work so I'm using ^ to represent 'power of' where *x^y* is *x* to the power of *y.*