Find the antiderivative of
e^(2lnx) + e^(2x)
recall that e^lnx = x. So, e^(2lnx) = (e^lnx)^2 = x^2
So, you just want
∫x^2 + e^(2x) = 1/3 x^3 + 1/2 e^(2x) + C
Well, this is an interesting problem. To find the antiderivative of e^(2lnx) + e^(2x), we can use some clever tricks.
First, let's simplify e^(2lnx). We know that ln(x) is the natural logarithm of x, so e^(2lnx) is really just x^2.
Now we have x^2 + e^(2x). Integrating x^2 is pretty straightforward - it's (1/3)x^3.
But what about e^(2x)? We don't have a handy formula for the antiderivative of e^(2x), so we have to get creative.
One way to approach this is to use integration by parts. Let's say u = e^(2x) and dv = dx. Then du = 2e^(2x) dx and v = x.
Using the integration by parts formula, which is ∫u dv = uv - ∫v du, we get:
∫e^(2x) dx = x*e^(2x) - ∫x*2e^(2x) dx
We can simplify the right side of the equation:
∫e^(2x) dx = x*e^(2x) - 2∫x*e^(2x) dx
Notice that we're essentially back to where we started, but with an extra x factor.
We can repeat this integration by parts process one more time to get rid of the x factor.
Let's say u = x and dv = e^(2x) dx. Then du = dx and v = (1/2)e^(2x).
Using the integration by parts formula again, we get:
∫x*e^(2x) dx = (1/2)x*e^(2x) - ∫(1/2)e^(2x) dx
Simplifying the right side, we have:
∫x*e^(2x) dx = (1/2)x*e^(2x) - (1/4)e^(2x)
Plugging this result back into our previous equation, we get:
∫e^(2x) dx = x*e^(2x) - 2*((1/2)x*e^(2x) - (1/4)e^(2x))
Simplifying further, we have:
∫e^(2x) dx = x*e^(2x) - x*e^(2x) + (1/2)e^(2x)
So the antiderivative of e^(2x) is (1/2)e^(2x).
Now we can put it all together:
∫e^(2lnx) + e^(2x) dx = ∫x^2 + e^(2x) dx
= (1/3)x^3 + (1/2)e^(2x) + C
And there you have it! The antiderivative of e^(2lnx) + e^(2x) is (1/3)x^3 + (1/2)e^(2x) + C, where C is the constant of integration.
To find the antiderivative of e^(2lnx) + e^(2x), we will break it down into two separate integrals:
1. Integral of e^(2lnx):
First, we can use a logarithmic property to rewrite e^(2lnx) as (e^(lnx))^2, which simplifies to x^2. So, the integral of e^(2lnx) with respect to x is equal to the integral of x^2 with respect to x.
Taking the integral of x^2, we use the power rule for integration, which states that the integral of x^n with respect to x is (x^(n+1))/(n+1). Applying this rule, we get:
∫ x^2 dx = (x^3)/3 + C1, where C1 is the constant of integration.
2. Integral of e^(2x):
The integral of e^(2x) with respect to x is straightforward. We can apply the power rule for integration again, keeping in mind that the coefficient of x is 2:
∫ e^(2x) dx = (1/2)e^(2x) + C2, where C2 is the constant of integration.
Combining both results, we get the antiderivative of the original function as:
∫ (e^(2lnx) + e^(2x)) dx = ∫ (x^2 + e^(2x)) dx = (x^3)/3 + (1/2)e^(2x) + C.
Therefore, the antiderivative of e^(2lnx) + e^(2x) is given by:
(x^3)/3 + (1/2)e^(2x) + C, where C is the constant of integration.
To find the antiderivative of the given expression, we can use the properties of logarithms and exponential functions.
Let's break down the given expression first:
e^(2ln(x)) + e^(2x)
We can simplify e^(2ln(x)) using the property of exponential functions:
e^(2ln(x)) = (e^(ln(x)))^2 = x^2
Now, the expression becomes:
x^2 + e^(2x)
To find the antiderivative, we will consider each term separately.
1. Antiderivative of x^2:
The antiderivative of x^n (where n is any real number except -1) is given by (x^(n+1))/(n+1). Applying this formula, the antiderivative of x^2 is:
∫ x^2 dx = x^3/3
2. Antiderivative of e^(2x):
The antiderivative of e^(kx) is given by (1/k) * e^(kx). In this case, k = 2, so:
∫ e^(2x) dx = (1/2) * e^(2x)
Now, combining the antiderivatives of both terms, we get:
∫ (x^2 + e^(2x)) dx = x^3/3 + (1/2) * e^(2x) + C
where C is the constant of integration.
Therefore, the antiderivative of e^(2ln(x)) + e^(2x) is x^3/3 + (1/2) * e^(2x) + C, where C is the constant of integration.