# Question Video: Finding the Indefinite Integral Using Standard Integrals Mathematics • Higher Education

Determine β«((7/4π₯) β 6π^(β2π₯)) dπ₯.

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### Video Transcript

Determine the integral of seven divided by four π₯ minus six times π to the power of negative two π₯ with respect to π₯.

The question is asking us to evaluate the integral of the difference between two functions. Our integrand is the difference between a reciprocal function and an exponential function. And we know how to integrate reciprocal functions and exponential functions. So weβll rewrite our integral of the difference between two functions as the difference between two integrals. So doing this to the integral given to us in the question, we get the integral of seven divided by four π₯ with respect to π₯ minus the integral of six times π to the power of negative two π₯ with respect to π₯. Before we evaluate each of these integrals separately, remember for a constant π the integral of π times π of π₯ with respect to π₯ is equal to π times the integral of π of π₯ with respect to π₯.

So in our first integral, we can take out our constant factor of seven over four. And in our second integral, we can take out our constant factor of six. This gives us seven over four times the integral of one over π₯ with respect to π₯ minus six times the integral of π to the power of negative two π₯ with respect to π₯. Weβre now ready to evaluate each of these integrals separately. Letβs start with the integral of the reciprocal function with respect to π₯. The integral of the reciprocal function with respect to π₯ is a standard integral which we should know. Itβs the natural logarithm of the absolute value of π₯ plus the constant of integration πΆ. So evaluating our first integral we get seven over four times the natural logarithm of π₯ plus the constant of integration weβll call πΆ one.

We now want to evaluate our second integral, which is the integral of π to the power of negative two π₯ with respect to π₯. This is the integral of an exponential function. And we know how to integrate exponential functions. For any constant π, where π is not equal to zero, the integral of π to the power of ππ₯ with respect to π₯ is equal to π to the power of ππ₯ divided by π plus the constant of integration πΆ. We need to divide by the coefficient of π₯ in our exponent. In this case, we see the coefficient of π₯ in our exponent is negative two. So our value of π is negative two. And this tells us the integral of π to the power of negative two π₯ with respect to π₯ is equal to π to the power of negative two π₯ divided by negative two plus our constant of integration weβll call πΆ two.

The next thing weβll do is simplify this expression by distributing our coefficients over our parentheses. Distributing seven over four over our first set of parentheses, we get seven over four times the natural logarithm of the absolute value of π₯ plus seven πΆ one divided by four. And distributing negative six over our second set of parentheses, we get three π to the power of negative two π₯ minus six πΆ two. And weβll do one last thing to simplify this expression. Since πΆ one and πΆ two are both constants of integration, seven πΆ one over four minus six πΆ two is just a constant. So weβll combine these into one constant, weβll call πΆ.

And this gives us our final answer. The integral of seven divided by four π₯ minus six times π to the power of negative two π₯ with respect to π₯ is equal to seven over four times the natural logarithm of the absolute value of π₯ plus three times π to the power of negative two π₯ plus πΆ.