Question

integration of 1/√(3x^2+5x+7)

Answer 1

Step-by-step explanation:

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Answer 2

Given :

A function f(x) is which is given as :

$f(x)=\frac{1}{\sqrt{3x^{2}+5x+7 } }$

To Find :

What is the integration of the given function f (x) i.e. :

$\int f(x).dx =?$

Solution :

Let $I$ be the integration of the function f (x) ,then $I$ will be :

$I=\int f(x).dx$

Taking √3 common from the denominator term we get :

$I=\frac{1}{\sqrt{3} } \int\frac{dx}{\sqrt{x^{2}+5x\frac{1}{3} +\frac{7}{3} } }$

Now we are adding and subtracting square of $\frac{5}{6}$ in the denominator term trying to make a perfect square ,so :

$I=\frac{1}{\sqrt{3} }\int\frac{dx}{\sqrt{x^{2}+2\times \frac{5}{6}x+(\frac{5}{6}) ^{2}-(\frac{5}{6}) ^{2} +\frac{7}{3} } }$

$I=\frac{1}{\sqrt{3} }\int\frac{dx}{\sqrt{(x+\frac{5}{6}) ^{2}+\frac{59}{36} } }$

Now we will use here the appropriate integration formula here which is :

$\frac{dx}{\sqrt{x^2 + a^2} }= \log|x + \sqrt{x^2 + a^2} |$

On using this formula we can write :

$I=\frac{1}{\sqrt{3} } \log[(x+\frac{5}{6})+\sqrt{(x+\frac{5}{6}) ^{2}+\frac{59}{36} }]$

$I=\frac{1}{\sqrt{3} } \log[(x +\frac{5}{6} )+\sqrt{x^{2}+\frac{5}{3}x+\frac{7}{3} }]+c$

So, the integration of  given function f (x) is $I=\frac{1}{\sqrt{3} } \log[(x +\frac{5}{6} )+\sqrt{x^{2}+\frac{5}{3}x+\frac{7}{3} }]+c$