Question:
Evaluate the following integrals:
$\int(3 x+4)^{2} d x$
Solution:
Given:
$\int(3 x+4)^{2} d x$
By applying,
$(a+b)^{2}=a^{2}+b^{2}+2 a b$
$\Rightarrow \int\left((3 x)^{2}+4^{2}+2 \times 3 x \times 4\right) d x$
$\Rightarrow \int\left(9 x^{2}+16+24 x\right) d x$
By Splitting, we get,
$\Rightarrow \int 9 x^{2} d x+\int 16 d x+\int 24 x d x$
$\Rightarrow 9 \int x^{2}+16 \int d x+24 \int x d x$
By applying,
$\int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}$
$\int \mathrm{kdx}=\mathrm{kx}+\mathrm{c}$
$\Rightarrow \frac{9 x^{2+1}}{2+1}+16 x+\frac{24 x^{1+1}}{1+1}+c$
$\Rightarrow \frac{9}{3} x^{3}+16 x+\frac{24}{2} x^{2}+c$
$\Rightarrow 3 x^{3}+16 x+12 x^{2}+c$