$\int _{4} ^{5} \dfrac {1} {5x-1} dx$

$\int _{0} ^{1} \dfrac {1} {2x-1} dx$

$\int _{ \infty } ^{ Â -\infty} \dfrac { \sin (x) } {1 + 3x ^ {2}} dx$

$\int _{1} ^{3} \ln (x-1) dx$

This **article aims**Â to find **improper integrals. **This article uses the concept of **improper integrals**; there are **two types**Â of improper integrals. Integrals having **limits of integrations are infinite**Â and are improper. When integrals become **infinite within the limits of integration**, integrals are improper.

**Improper integrals**Â can be represented as follows:

\[ \int _{ -\infty} ^{ \infty} f(x) dx\]

**Expert Answer**

An**Â improper integral**Â is a definite integral **with one or both limits**Â at infinity or an**Â integrand that approaches infinity**Â at one or more points in the range of **integration.**

**Part 1**

Let’s solve for each integral.

\[\int _{4} ^{5} \dfrac {1} {5x-1} dx\]

since none of the **limits is equal to infinity**, we check the **integrand.**

As we know that

\[ \dfrac {1}{x} = \ln |x|\]

\[\int _{4} ^{5} \dfrac {1} {5x-1} dx = \dfrac {1} {5} \ln |5x-1| |_ {4} ^{5}\]

\[ = \dfrac {1} {5} ( \ln |24| – \ln |19|)\]

The integrand is not **approaching infinity,**Â so the given**Â integral is not improper.**

**Part 2**

\[\int _{0} ^{1} \dfrac {1} {2x-1} dx \]

**Plugging**Â $x=\dfrac {1} {2}$, the **integral approaches**Â to $\infty$ and $ x = \dfrac {1}{2} $ is also in the range of integration $[ 0 , 1 ]$.

So the **given integral is improper.**

**Part 3**

\[ \int _{ \infty } ^{ Â -\infty} \dfrac { \sin (x) } {1 + 3x ^ {2}} dx \]

As seen clearly, the limits of **integrations are infinity**, so by definition, the given **integral is improper.**

**Part 4**

\[ \int _{1} ^{3} \ln (x-1) dx \]

At $x = 1$, the **integral becomes undefined, **and this value is also in the **range of integration.**

So it is also an **improper fraction.**

**Numerical Result**

$\int _{4} ^{5} \dfrac {1} {5x-1} dx$ is not an **improper integral.**

$\int _{0} ^{1} \dfrac {1} {2x-1} dx$ is an **improper integral.**

$\int _{ \infty } ^{ Â -\infty} \dfrac { \sin (x) } {1 + 3x ^ {2}} dx$ is an **improper integral.**

$\int _{1} ^{3} \ln (x-1) dx$ is an **improper integral.**

**Example**

**Which of the following integrals is improper?**

$\int _{- \infty} ^{ \infty } \dfrac {1} {6x+1} dx $

$\int _{5} ^{9} \dfrac {1} {5x+1} dx$

**Solution**

**Part a**

\[ \int _{- \infty} ^{ \infty } \dfrac {1} { 6x+1} dx \]

As seen clearly, the limits of **integrations are infinity**, so by definition, the given **integral is improper.**

**Part b**

Since none of the **limits is equal to infinity**, we check the **integrand.**

As we know:

\[ \dfrac {1}{x} = \ln |x|\]

\[\int _{5} ^{9} \dfrac {1} {5x+1} dx = \dfrac {1} {5} \ln |5x+1| |_ {5} ^{9}\]

\[ = \dfrac {1} {5} ( \ln |46| – \ln |26|)\]

As the integrand is not **approaching infinity,**Â the given**Â integral is not improper.**