Question:
Evaluate the following integrals:
$\int \frac{4 x+3}{\sqrt{2 x^{2}+3 x+1}} d x$
Solution:
Assume, $2 x^{2}+3 x+1=t$
$d\left(x^{2}+x+1\right)=d t$
$(4 x+3) d x=d t$
Substituting $t$ and dt in above equation we get
$\Rightarrow \int \frac{1}{\sqrt{t}} \mathrm{dt}$
$\Rightarrow \int \mathrm{t}^{-1 \backslash 2} \cdot \mathrm{dt}$
$\Rightarrow 2 \mathrm{t}^{1 \backslash 2}+\mathrm{c}$
But $t=2 x^{2}+3 x+1$
$\Rightarrow 2\left(2 x^{2}+3 x+1\right)^{1 / 2}+c$
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