Contents

**article aims**to find

**improper integrals.**This article uses the concept of

**improper integrals**; there are

**two types**of improper integrals. Integrals having

**limits of integrations are infinite**and are improper. When integrals become

**infinite within the limits of integration**, integrals are improper.

**Improper integrals**can be represented as follows: \[ \int _{ -\infty} ^{ \infty} f(x) dx\]

**Expert Answer**

An**improper integral**is a definite integral

**with one or both limits**at infinity or an

**integrand that approaches infinity**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,**so the given

**integral is not improper.**

**Part 2**

\[\int _{0} ^{1} \dfrac {1} {2x-1} dx \]
**Plugging**$x=\dfrac {1} {2}$, the

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

**given integral is improper.**

**Part 3**

\[ \int _{ \infty } ^{ -\infty} \dfrac { \sin (x) } {1 + 3x ^ {2}} dx \]
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.**$\int _{ \infty } ^{ -\infty} \dfrac { \sin (x) } {1 + 3x ^ {2}} dx$ 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,**the given

**integral is not improper.**

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