# { \int (12x+6)(3x^2+3x+5)^7 dx= ? }

## Question:

{eq}\int (12x+6)(3x^2+3x+5)^7 dx= ? {/eq}

## Integration By Substitution:

There are some integrals that are difficult to solve directly. If we observe them closely, we can find ways to simplify integrals into more familiar forms.

Substitution is the easiest method to solve the difficult integral. In this method, we define a new function and integrate the integral over that instead of over {eq}x {/eq}. This method is applicable only if we have an integral of the form {eq}\int f(p(x)) p'(x) dx {/eq}

We have to solve the integral

{eq}\int (12x+6)(3x^2+3x+5)^7 dx {/eq}

We solve the integral by using the substitution method.

We use the substitution {eq}3x^2+3x+5=t {/eq} to solve the integral.

Differentiate both sides of the substitution with respect to {eq}x {/eq}

{eq}\begin{align} 3x^2+3x+5 &=t\\ \dfrac{\mathrm{d} }{\mathrm{d} x}(3x^2+3x+5) &=\dfrac{\mathrm{d} t}{\mathrm{d} x}\\ 3(2x)+3(1)+0 &=\dfrac{\mathrm{d} t}{\mathrm{d} x}\\ (6x+3)dx &=dt \end{align} {/eq}

Applying this substitution to the integral, we have:

{eq}\begin{align} \int (12x+6)(3x^2+3x+5)^7 dx &=\int 2(6x+3)(3x^2+3x+5)^7 dx\\ &=\int 2(t)^7dt\\ &=2\int t^7dt\\ &=2\left ( \dfrac{t^8}{8} \right )+C\\ &=\dfrac{t^8}{4}+C \end{align} {/eq}

Reversing the substitution, we have:

{eq}\begin{align} \int (12x+6)(3x^2+3x+5)^7 dx &=\dfrac{t^8}{4}+C\\ &=\dfrac{1}{4}(3x^2+3x+5)^8+C \end{align} {/eq}

{eq}\color{blue}{\int (12x+6)(3x^2+3x+5)^7 dx=\dfrac{1}{4}(3x^2+3x+5)^8+C} {/eq} 