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

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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 6n 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.

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