Indefinite Integrals – Techniques with Exponentials & Logs

An antiderivative acts as the reverse operation of a derivative. By definition, a function F is an antiderivative of f on an interval if F'(x) = f(x) for all x in that interval. This means that if you differentiate the answer (the antiderivative), you must get back the original function (the integrand).

To find the antiderivative of a power function like 5x⁴, we use the Power Rule for Integration, which states that the integral of xⁿ is xⁿ⁺¹/(n+1). Here, the power n is 4. We add 1 to the exponent to get 5, and then divide the coefficient 5 by the new exponent 5. This results in (5/5)x⁵, which simplifies to x⁵. We add the constant of integration C to complete the indefinite integral.

We apply the Power Rule for Integration here. The current exponent is 2/3. First, we add 1 to the exponent: 2/3 + 1 equals 5/3. Next, we divide the term by this new exponent. Dividing by 5/3 is the same as multiplying by its reciprocal, 3/5. Therefore, the result is (3/5)x^(5/3) + C.

We can factor the constant 4 out of the integral, leaving us with 4 times the Integral of (1/x) dx. The standard antiderivative of 1/x is the natural logarithm of the absolute value of x, written as ln|x|. Multiplying this by the constant 4 gives us 4 ln|x| + C.

We integrate each term separately. The antiderivative of cos(x) is sin(x) because the derivative of sin(x) is cos(x). For the second term, -2sin(x), we recall that the derivative of cos(x) is -sin(x). Therefore, the antiderivative of -sin(x) is cos(x). Multiplying by the constant 2, the antiderivative of -2sin(x) is 2cos(x). Combining these gives sin(x) + 2cos(x) + C.

This requires finding the antiderivative of an exponential function with a linear composite power. We know the derivative of e^(5x) is 5e^(5x) due to the chain rule. To reverse this, when we integrate e^(kx), we must divide by the constant k. Here k is 5, so the integral is (1/5)e^(5x) + C.

When we find an indefinite integral, we are looking for a family of functions whose derivative is the integrand. Because the derivative of any constant number is zero, functions like x², x² + 10, and x² - 5 all have the same derivative, 2x. The "+ C" represents any possible constant value, ensuring we have captured the general solution rather than just one specific case.

We can solve this by recognizing it as a power function with a linear inner function (x+3) where the slope is 1. Using the reverse power rule, we add 1 to the exponent to get 3, and divide by the new exponent 3. Since the derivative of the inside (x+3) is just 1, no further adjustment is needed. Alternatively, you could expand the term to x² + 6x + 9 and integrate term by term to get x³/3 + 3x² + 9x + C, which is algebraically equivalent to (⅓)(x+3)³ + C (the constants merge).

Before integrating, we must simplify the integrand algebraically. We divide each term in the numerator by the denominator x². This gives us x + 5 - 4x^(-2). Now we integrate term by term. The integral of x is (x²)/2. The integral of 5 is 5x. The integral of -4x^(-2) uses the power rule: add 1 to the exponent (-2+1 = -1) and divide by the new exponent (-1). This results in -4(x^(-1))/(-1), which simplifies to +4/x. Combining these yields (x²)/2 + 5x + 4/x + C.

We are looking for a function whose derivative is sin(2x). We know the derivative of cos(u) is -sin(u) * u'. If we guess -cos(2x), the derivative would be -(-sin(2x) * 2) = 2sin(2x). This is double what we want. To correct this, we multiply our guess by 1/2. Therefore, the correct antiderivative is -(½)cos(2x) + C.

The integrand is an exponential function composed with a linear function (3x - 2). The rule for integrating e^(ax+b) is (1/a)e^(ax+b) + C. Here, a = 3. Therefore, we divide by 3 to compensate for the chain rule that would occur during differentiation. The result is (⅓)e^(3x - 2) + C.

This is a composition involving the natural logarithm rule. The integral of 1/u is ln|u|, but here u = 2x + 5. Because the inner function is linear with a coefficient of 2, we must divide the result by 2 to reverse the chain rule. Thus, the integral is (½)ln|2x + 5| + C.

While all continuous functions have antiderivatives (in the sense of area accumulation), not all can be expressed using a finite combination of polynomials, exponentials, logarithms, and trigonometric functions. The function e^(x²) is a classic example of a function that is integrable (it has an area) but possesses no elementary closed-form antiderivative.

We do not have a direct rule for tan²(x), so we use a trigonometric identity. We know that tan²(x) + 1 = sec²(x), which means tan²(x) = sec²(x) - 1. We can now integrate the rewritten expression: Integral of (sec²(x) - 1) dx. The antiderivative of sec²(x) is tan(x), and the antiderivative of -1 is -x. Therefore, the solution is tan(x) - x + C.

First, rewrite the integrand using exponents: (3x)^(-½). This is a linear composition where the inner slope is 3. Using the power rule, we add 1 to the exponent (-1/2 + 1 = 1/2) and divide by the new exponent (½), which is equivalent to multiplying by 2. This gives 2(3x)^(½). However, because of the inner function 3x, we must also divide by the derivative of the inside, which is 3. Combining these steps results in (⅔)(3x)^(½) + C, or (⅔)√(3x) + C.