In this article, we will see the integration rules to be followed for solving an integral of the type e^{x} [f(x) + f ’(x)], where f ’(x) is the derivative of f(x). We will use integration by parts and some other integration rules to solve these equations.

**Browse more Topics Under Integrals**

- Fundamental Theorem of Calculus
- Introduction to Integration
- Properties of Indefinite Integrals
- Properties of Definite Integrals
- Definite Integral as a Limit of a Sum
- Integration by Partial Fractions
- Integration by Parts
- Integration by Substitutions
- Integral of Some Particular Functions
- Integral of the Type e^x[f(x) + f'(x)]dx

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## Integral of the Type e^x[f(x) + f ‘(x)]dx

To begin with, let’s say

I = ∫ e^{x} [f(x) + f ’(x)] dx

Opening the brackets, we get,

I = ∫ e^{x} f(x) dx + ∫ e^{x} f ’(x) dx = I_{1} + ∫ e^{x} f ’(x) dx … (1)

Where, I_{1} = ∫ e^{x} f(x) dx

To solve I_{1}, we will use integration by parts. Let the first function = f_{1}(x) = f(x) and the second function = g_{1}(x) = e^{x}. Therefore,

I_{1} = f(x) ∫ e^{x} dx – ∫ [df(x)/dx ∫ e^{x} dx] dx

Or, I_{1} = e^{x} f(x) – ∫ e^{x} f ’(x) dx + C

Substituting the value of I1 in equation (1), we get

I = e^{x} f(x) – ∫ e^{x} f ’(x) dx + ∫ e^{x} f ’(x) dx + C = e^{x} f(x) + C

Thus, ∫ e^{x} [f(x) + f ’(x)] dx = e^{x} f(x) + C … (2)

Let’s look at an example to understand it well.

*(Source: blue diamond)*

### Integration Rules – Example 1

Find ∫ e^{x} {tan^{–1} x + [1 / (1 + x^{2})]} dx

Solution: Let, f(x) = tan^{–1} x.

Therefore, f ’(x) = df(x)/dx = d(tan^{–1} x)/ dx = 1 / (1 + x^{2})

Hence, the integrand is of the form: e^{x} [f(x) + f ’(x)]. Therefore, using equation (2), we get

∫ e^{x} {tan^{–1} x + [1 / (1 + x^{2})]} dx = e^{x} tan^{–1} x + C

### Integration rules – Example 2

Find ∫ [e^{x} (x^{2} + 1) / (x + 1)^{2}] dx

Solution: We have,

I = ∫ [e^{x} (x^{2} + 1) / (x + 1)^{2}] dx = ∫ [e^{x} (x^{2} + 1 + 1 – 1) / (x + 1)^{2}] dx

= ∫ [e^{x} {(x^{2} – 1) / (x + 1)^{2}] + 2 / (x + 1)^{2}}] dx

Therefore, I = ∫ [e^{x} {(x – 1) / (x + 1)] + 2 / (x + 1)^{2}}] dx

Now, let f(x) = (x – 1) / (x + 1). Therefore,

f ’(x) = df(x) / dx = d [(x – 1) / (x + 1)] / dx = 2 / (x + 1)^{2}Hence, the integrand is of the form: e^{x} [f(x) + f ’(x)]. Therefore, using equation (2), we get

∫ [e^{x} (x^{2} + 1) / (x + 1)^{2}] dx = e^{x} (x – 1) / (x + 1) + C

## More Solved Examples for You

**Question 1: Find ∫ e ^{x} (sin x + cos x) dx**

**Answer :**f ’(x) = df(x)/dx = d sin x / dx = cos x.

Hence, the integrand is of the form: e^{x} [f(x) + f ’(x)]. Therefore, using equation (2), we get

∫ e^{x} (sin x + cos x) dx = e^{x} sin x + C

**Question 2: Find ∫ e ^{x} [(1 / x) – (1 / x^{2})] dx**

**Answer :** Let, f(x) = 1/x. Therefore,

f ’(x) = df(x)/dx = d(1/x)/dx = 1/x^{2}.

Hence, the integrand is of the form: e^{x} [f(x) + f ’(x)]. Therefore, using equation (2), we get

∫ e^{x} [(1 / x) – (1 / x^{2})] dx = e^{x} (1/x) + C = e^{x}/x + C

**Solved Questions for You**

**Question 1: What are integration and differentiation?**

**Answer:** Differentiation refers to the act of finding the rate of change of the gradient/slope of any function whereas integration refers to the area under the curve of function with regards to the x-axis.

**Question 2: What is the product rule of integration?**

**Answer:** The Product Rule of integration allows us to integrate the product of two functions. For instance, through a series of mathematical somersaults, we can turn any equation into a formula which is useful for integrating.

**Question 3: Who introduced integration?**

**Answer:** Gottfried Wilhelm Leibniz formulated the principle of integration along with Isaac Newton. It originated in the late 17^{th} century where they considered the integral as an infinite sum of rectangles of infinitesimal width.

**Question 4: What is called integration?**

**Answer:** Integration basically refers to bringing together and uniting of things. For instance, the integration of two or more economies, cultures, religions and so on. However, in math, integration refers to a concept of calculus, which is the act of finding integrals.