Question Video: Finding the Integration of a Function Using Integration by Substitution | Nagwa Question Video: Finding the Integration of a Function Using Integration by Substitution | Nagwa

# Question Video: Finding the Integration of a Function Using Integration by Substitution Mathematics • Third Year of Secondary School

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Determine β« π₯β΅(π₯βΆ + 9)β· dπ₯.

03:07

### Video Transcript

Determine the integral of π₯ to the fifth power multiplied by π₯ to the sixth power plus nine all to the seventh power with respect to π₯

This is not a polynomial thatβs nice to integrate using our standard rules for finding the antiderivative. And we certainly do not want to distribute our parentheses and find the antiderivative for each term. Instead, we spot that the integral is set up in this form. We have some function π of π₯ and its derivative π prime of π₯. If we look carefully, we see that π₯ to the fifth power is a scalar multiple of the derivative of π₯ to the power of six plus nine. And this means we can use integration by substitution to evaluate our indefinite integral.

The substitution rule says that if π’ is equal to π of π₯ is a differentiable function whose range is some interval π and π is continuous on this interval, then the integral of π of π of π₯ multiplied by π prime of π₯ with respect to π₯ is equal to the integral of π of π’ with respect to π’. Weβre going to let π’ be equal to the function that we originally defined π of π₯. So, π’ is equal to π₯ to the sixth power plus nine.

Now, this is great, as when we differentiate this function π’ with respect to π₯, we see that dπ’ by dπ₯ is equal to six π₯ to the fifth power. In integration by substitution, we think of the dπ’ and dπ₯ as differentials. And we can alternatively write this as dπ’ equals six π₯ to the fifth power dπ₯. Notice that whilst dπ’ by dπ₯ is definitely not a fraction, we do treat it a little like one in this process. We divide through by six, and we see that a sixth dπ’ is equal to π₯ to the fifth power dπ₯.

And now, letβs look back to our original integral. We see that we can replace π₯ to the fifth power dπ₯ with a sixth dπ’. And we replace π₯ to the sixth power plus nine with π’. And we now see that the integral weβre evaluating is the integral of a sixth of π’ to the seventh power dπ’. If we so choose, we can take this factor of a sixth outside of the integral sign, and then weβre evaluating a sixth of the integral of π’ to the seventh power with respect to π’.

The integral of π’ to the seventh power is π’ to the eighth power divided by eight plus, since it is an indefinite integral, π the constant of integration. We distribute our parentheses. And we see that the integral is equal to one over 48 times π’ to the eighth power plus πΆ. And notice Iβve chosen this to be a capital πΆ because our original constant of integration has been multiplied by one-sixth.

But, remember, we were originally looking to evaluate our integral in terms of π₯. So, we look to our original definition of π’. And we said that π’ was equal to π₯ to the sixth power plus nine. And we then see that our integral is equal to one over 48 times π₯ to the sixth power plus nine to the eighth power plus πΆ.

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