# 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.

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

{eq}\displaystyle I =...

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