Solve integration problem. Involving inverse function, log function
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Today we'll solve an integration problem involving logarithmic functions. We want to find the integral of natural log of x squared plus one. We'll use integration by parts, where u equals natural log of x squared plus one, and dv equals dx.
Now we calculate du and v. The derivative of natural log of x squared plus one is two x divided by x squared plus one, so du equals this expression times dx. For v, we simply integrate dx to get x. Applying integration by parts, we get x times natural log of x squared plus one minus the integral of two x squared over x squared plus one dx.
We need to simplify the fraction two x squared over x squared plus one. We can rewrite this as two times x squared plus one minus two, all over x squared plus one, which simplifies to two minus two over x squared plus one. When we integrate this, we get two x minus two arctan x plus C, using the standard integral form for one over x squared plus one.
Now we substitute our result back into the integration by parts formula. We get x times natural log of x squared plus one minus the quantity two x minus two arctan x. Simplifying this gives us our final answer: x natural log of x squared plus one minus two x plus two arctan x plus C. This solution beautifully combines both logarithmic and inverse trigonometric functions.
To summarize what we've learned: Integration by parts is the key technique for integrating logarithmic functions. We used algebraic manipulation to simplify complex fractions and applied standard integral forms for inverse trigonometric functions. Our final solution elegantly combines both logarithmic and inverse trigonometric functions, demonstrating how different mathematical concepts work together in calculus.