# Find the integral \int \frac {16}{1+64x^2} dx

## Question:

Find the integral {eq}\int \frac {16}{1+64x^2} dx {/eq}

## Integration by Substitution

Integration by substitution is the way to integrate a complicated integral. In this method, a certain group of variables or a variable is replaced by a single variable to make integration easy after integration again assumed variable is replaced with the original one.

Given data

• The integral is: {eq}\int {\dfrac{{16}}{{1 + 64{x^2}}}dx} {/eq}

Rewrite the given integral function as,

{eq}I = \int {\dfrac{{16}}{{1 + {{\left( {8x} \right)}^2}}}dx} {/eq}

Let us assume that {eq}t = 8x {/eq}.

Differentiate the above equation.

{eq}\begin{align*} \dfrac{{dt}}{{dx}} &= 8\\ \dfrac{{dt}}{8} &= dx \end{align*} {/eq}

Substitute the known values in original function,

{eq}\begin{align*} I &= \int {\dfrac{{16}}{{1 + {{\left( t \right)}^2}}}\dfrac{{dt}}{8}} \\ &= 2\int {\dfrac{{dt}}{{1 + {{\left( t \right)}^2}}}} \\ &= 2{\tan ^{ - 1}}\left( t \right) + k \end{align*} {/eq}

Here, constant is {eq}k {/eq}.

Substitute the known values,

{eq}I = 2{\tan ^{ - 1}}\left( {8x} \right) + k {/eq}

Thus, the solution of the integral is {eq}2{\tan ^{ - 1}}\left( {8x} \right) + k {/eq}.