# Evaluate the integral: Integral 3(squareroot{(2x + 5)^2 - 9}) / (4x + 10) dx using the formula,...

## Question:

Evaluate the integral:

{eq}\displaystyle \int \frac{\displaystyle 3\; \sqrt{(2x + 5)^2 - 9}}{\displaystyle (4x + 10)}dx {/eq}

using the formula,

{eq}\displaystyle \int \frac{\displaystyle \sqrt {u^2 - a^2}}{\displaystyle u} du = {\sqrt{u^2 - a^2}} - a \sec^{-1}\bigg (\frac{\displaystyle |u|}{\displaystyle a}\bigg ) + C. {/eq}

## Indefinite Integral:

Given any function, {eq}f(x) {/eq}, an anti-derivative of {eq}f(x) {/eq} is any function {eq}F(x) {/eq} such that {eq}F'(x) = f(x). {/eq}

If {eq}F(x) {/eq} is anti-derivative of {eq}f(x) {/eq}, then the most general anti-derivative of {eq}f(x) {/eq} is called an indefinite integral and denoted,

{eq}\int {f(x)dx = F(x) + c} {/eq}, where c is constant of integration.

We use the following formula in the given problem:

{eq}\displaystyle \int \frac{\displaystyle \sqrt {u^2 - a^2}}{\displaystyle u} du = {\sqrt{u^2 - a^2}} - a \sec^{-1}\bigg (\frac{\displaystyle |u|}{\displaystyle a}\bigg ) + c, {/eq},

where {eq}c {/eq} is arbitrary constant.

## Answer and Explanation:

Let {eq}I = \displaystyle \int \frac{\displaystyle 3\; \sqrt{(2x + 5)^2 - 9}}{\displaystyle (4x + 10)}dx, {/eq} then

{eq}\displaystyle I =...

See full answer below.

Become a Study.com member to unlock this answer! Create your account

View this answerLet {eq}I = \displaystyle \int \frac{\displaystyle 3\; \sqrt{(2x + 5)^2 - 9}}{\displaystyle (4x + 10)}dx, {/eq} then

{eq}\displaystyle I = \frac{3}{2}\int {\frac{{\sqrt {{{\left( {2x + 5} \right)}^2} - 9} }}{{\left( {2x + 5} \right)}}} dx, {/eq}

Substitute {eq}\displaystyle 2x + 5 = u,2dx = du, {/eq}

{eq}\displaystyle \eqalign{ & I = \frac{3}{2}\int {\frac{{\sqrt {{u^2} - 9} }}{u}} \frac{{du}}{2} \cr & I = \frac{3}{4}\int {\frac{{\sqrt {{u^2} - 9} }}{u}} du, \cr} {/eq}

Now applying the above given formula,

{eq}\displaystyle I = \frac{3}{4}\left[ {\sqrt {{u^2} - 9} - 3{{\sec }^{ - 1}}\left( {\frac{{\left| u \right|}}{3}} \right)} \right] + c, {/eq}

Substituting back the value of {eq}u, {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from Calculus: Tutoring Solution

Chapter 7 / Lesson 14