Help please integral of sqrt (x^2+6x) dx

1 answer:

3 0

The integral of (x^2 + 6x)dx is 1/3x^3 + 3x^2 + c.
Because this is not an integration with specific bounds, you must include a constant at the end. In general, to integrate, add 1 to the exponent of x and then whatever number is the exponent of x, divided the number in front of x by that.

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