 Welcome to the GDSU Calculus Screencasts. In this episode, we'll consider a specific example of integration by substitution. Recall that integration by substitution is developed by undoing the chain rule for differentiation. The chain rule says that to differentiate a composite, f of g of x, we take f prime of g of x and multiply by g prime of x. And since anti-differentiation is the reverse of differentiation, we obtain the integral formula. Integral of f prime of g of x times g prime of x dx is f of g of x plus c. This is our substitution formula. And c here is an arbitrary constant that gives us all of the anti-derivatives of the integrand. Now in this screencast, we'll apply this technique of substitution to find an anti-derivative of 5 over 1 plus 25x squared. To apply the method of substitution, we try to recognize our integrand as a composite. So pause the video for a moment and see if you can rewrite the integrand as a composite of functions, and then figure out what substitution we might try. And as a hint, notice that 25x squared is the square of 5x. Resume the video when you're ready. Note that we can rewrite the integrand so that it has the form 5 over 1 plus the quantity 5x squared. And 1 plus the quantity 5x squared is the composite of 1 plus x squared with 5x, with 5x being the innermost function. So a substitution that we might try would be u equals 5x. Now go ahead and pause the video and try making the substitution into that integral to complete the integration. Resume the video when you're ready. Now we're trying the substitution u equal 5x. If we set u equal to 5x, that makes du 5 dx. Now when we make this substitution, notice that the 5x in red is our u, and the 5 dx in blue is our du. So we reduce the integral of 5 over 1 plus 5x squared dx to the integral of 1 over 1 plus u squared du. Now this last integral is one we should recognize as an arc tangent. So evaluating that last integral gives us the integral of 5 over 1 plus 5x squared dx is the integral of 1 over 1 plus u squared du is the arc tangent of u plus a constant. And to complete the integration, since we started with a function of x, we should wind up with a function of x. So we replace u with 5x to obtain the integral of 5 over 1 plus 25x squared dx is the arc tangent of 5x plus a constant. And we're done. Now one final note. With this integration process, we can always check our work by differentiating our result to see if we get the original integrand back. So in this case, we would differentiate our result arc tangent of 5x. And because we have a composite, we'd use the chain rule. Differentiating the arc tangent gives us 1 over 1 plus the square of what's inside. So we get 1 over 1 plus 5x squared. Then we have to multiply by the derivative of what's inside. That's 5. And that's exactly the integrand we had. So we've successfully found all the antiderivatives of 5 over 1 plus 25x squared. That concludes this screencast on integration by substitution. And we hope you'll come back and visit again in the future.