Final Answer
$400t-800\ln\left(\frac{2\sqrt{6}}{\sqrt{24+t^2}}\right)+C_0$
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Step-by-step Solution
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1
The integral of a function times a constant ($400$) is equal to the constant times the integral of the function
$400\int\left(1+\frac{2t}{24+t^2}\right)dt$
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$400\int\left(1+\frac{2t}{24+t^2}\right)dt$
Learn how to solve integral calculus problems step by step online. Find the integral int(400(1+(2t)/(24+t^2)))dt. The integral of a function times a constant (400) is equal to the constant times the integral of the function. Simplify the expression inside the integral. The integral 400\int1dt results in: 400t. The integral 800\int\frac{t}{24+t^2}dt results in: -800\ln\left(\frac{2\sqrt{6}}{\sqrt{24+t^2}}\right).
Final Answer
$400t-800\ln\left(\frac{2\sqrt{6}}{\sqrt{24+t^2}}\right)+C_0$