Question

Evaluate the following integrals: $\int {\sqrt {2{x^2} + 3x + 4} dx} $.
A) $\left( {4x + 3} \right)\sqrt {2{x^2} + 3x + 4} + \dfrac{{23\sqrt 2 }}{{32}}\log \left| {\left( {x + \dfrac{3}{4}} \right) + \dfrac{1}{{\sqrt 2 }}.\sqrt {2{x^2} + 3x + 4} } \right| + C$
B) $\dfrac{1}{8}\left( {4x + 3} \right)\sqrt {2{x^2} + 3x + 4} + \dfrac{{23\sqrt 2 }}{{32}}\log \left| {\left( {x + \dfrac{3}{4}} \right) + \dfrac{1}{{\sqrt 2 }}.\sqrt {2{x^2} + 3x + 4} } \right| + C$
C) $\dfrac{1}{8}\left( {4x + 3} \right)\sqrt {2{x^2} + 3x + 4} + \dfrac{{\sqrt 2 }}{{32}}\log \left| {\left( {x + \dfrac{3}{4}} \right) + \dfrac{1}{{\sqrt 2 }}.\sqrt {2{x^2} + 3x + 4} } \right| + C$
D) None of these

Answer 1

Hint:
We have asked in the question to evaluate $\int {\sqrt {2{x^2} + 3x + 4} dx} $.
Then after, we will take out 2 common from the above equation. Then, we will make the equation a perfect square equation for that we will add and subtract $\dfrac{9}{{16}}$.
Then, we will apply $\int {\sqrt {{x^2} + {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}} + \dfrac{{{a^2}}}{2}\log \left( {x + \sqrt {{x^2} + {a^2}} } \right) + C$ on the given equation.
Finally, after solving we will get the required answer.

Complete step by step solution:
It is given in the question to evaluate $\int {\sqrt {2{x^2} + 3x + 4} dx} $ .
Let, $I = \int {\sqrt {2{x^2} + 3x + 4} dx} $
Now, take out 2 common from the above equation, we get,
 $I = \int {\sqrt 2 \sqrt {{x^2} + \dfrac{{3x}}{2} + 2} dx} $
To make \[{x^2} + \dfrac{{3x}}{2} + 2\] a perfect square equation we will add and subtract $\dfrac{9}{{16}}$ .
 $I = \int {\sqrt 2 \sqrt {{x^2} + \dfrac{{3x}}{2} + \dfrac{9}{{16}} - \dfrac{9}{{16}} + 2} dx} $
 $I = \int {\sqrt 2 \sqrt {\left( {{x^2} + \dfrac{{3x}}{2} + \dfrac{9}{{16}}} \right) - \dfrac{9}{{16}} + 2} dx} $
 $I = \int {\sqrt 2 \sqrt {{{\left( {x + \dfrac{3}{4}} \right)}^2} + \dfrac{{32 - 9}}{{16}}} dx} $
 $I = \int {\sqrt 2 \sqrt {{{\left( {x + \dfrac{3}{4}} \right)}^2} + \dfrac{{23}}{{16}}} dx} $
 \[I = \int {\sqrt 2 \sqrt {{{\left( {x + \dfrac{3}{4}} \right)}^2} + {{\left( {\dfrac{{\sqrt {23} }}{4}} \right)}^2}} dx} \]
Now, the above equation is in the form of $\int {\sqrt {{x^2} + {a^2}} } dx$ where we have $x + \dfrac{3}{4}$ in place of $'x'$and $\dfrac{{\sqrt {23} }}{4}$ in place of $'a'.$
Since, $\int {\sqrt {{x^2} + {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}} + \dfrac{{{a^2}}}{2}\log \left( {x + \sqrt {{x^2} + {a^2}} } \right) + C$ .
Now, apply $\int {\sqrt {{x^2} + {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}} + \dfrac{{{a^2}}}{2}\log \left( {x + \sqrt {{x^2} + {a^2}} } \right) + C$ in the equation \[\int {\sqrt 2 \sqrt {{{\left( {x + \dfrac{3}{4}} \right)}^2} + {{\left( {\dfrac{{\sqrt {23} }}{4}} \right)}^2}} dx} \] , we get,

 \[I = \sqrt 2 \left[ {\dfrac{{x + \dfrac{3}{4}}}{2}\sqrt {{{\left( {x + \dfrac{3}{4}} \right)}^2} + \dfrac{{23}}{{16}}} + \dfrac{{\dfrac{{23}}{{16}}}}{2}\log \left( {\left( {x + \dfrac{3}{4}} \right) + \sqrt {{{\left( {x + \dfrac{3}{4}} \right)}^2} + \dfrac{{23}}{{16}}} } \right)} \right] + C\]
Now, apply ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ on $\left( {x + \dfrac{3}{4}} \right)$
 \[I = \sqrt 2 \left[ {\dfrac{{4x + 3}}{8}\sqrt {{x^2} + 2 \times x \times \dfrac{3}{4} + \dfrac{9}{{16}} + \dfrac{{23}}{{16}}} + \dfrac{{23}}{{32}}\log \left( {\left( {x + \dfrac{3}{4}} \right) + \sqrt {{x^2} + 2 \times x \times \dfrac{3}{4} + \dfrac{9}{{16}} + \dfrac{{23}}{{16}}} } \right)} \right] + C\]
 \[I = \sqrt 2 \left[ {\dfrac{{4x + 3}}{8}\sqrt {{x^2} + \dfrac{{3x}}{2} + \dfrac{{32}}{{16}}} + \dfrac{{23}}{{32}}\log \left( {\left( {x + \dfrac{3}{4}} \right) + \sqrt {{x^2} + \dfrac{{3x}}{2} + \dfrac{{32}}{{16}}} } \right)} \right] + C\]
 \[I = \sqrt 2 \left[ {\dfrac{{4x + 3}}{8}\sqrt {{x^2} + \dfrac{{3x}}{2} + 2} + \dfrac{{23}}{{32}}\log \left( {\left( {x + \dfrac{3}{4}} \right) + \sqrt {{x^2} + \dfrac{{3x}}{2} + 2} } \right)} \right] + C\]
 \[I = \sqrt 2 \left[ {\left( {\dfrac{{4x + 3}}{8}} \right)\dfrac{{\sqrt {2{x^2} + 3x + 4} }}{{\sqrt 2 }} + \dfrac{{23\sqrt 2 }}{{32}}\log \left( {\left( {x + \dfrac{3}{4}} \right) + \dfrac{{\sqrt {2{x^2} + 3x + 4} }}{{\sqrt 2 }}} \right)} \right] + C\]
 \[\therefore I = \left( {\dfrac{{4x + 3}}{8}} \right)\sqrt {2{x^2} + 3x + 4} + \dfrac{{23\sqrt 2 }}{{32}}\log \left[ {\left( {x + \dfrac{3}{4}} \right) + \dfrac{{\sqrt {2{x^2} + 3x + 4} }}{{\sqrt 2 }}} \right] + C\]
Hence, $\int {\sqrt {2{x^2} + 3x + 4} dx} = \left( {\dfrac{{4x + 3}}{8}} \right)\sqrt {2{x^2} + 3x + 4} + \dfrac{{23\sqrt 2 }}{{32}}\log \left[ {\left( {x + \dfrac{3}{4}} \right) + \dfrac{{\sqrt {2{x^2} + 3x + 4} }}{{\sqrt 2 }}} \right] + C$

Therefore, option (B) is correct.

Note:
Some properties of integration:
1) $\int {\dfrac{{dx}}{{\sqrt {{x^2} + {a^2}} }}} = \ln \left( {x + \sqrt {{x^2} + {a^2}} } \right) + C$
2) $\int {\dfrac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} = \ln \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$
3) $\int {\sqrt {{a^2} - {x^2}} } dx = \dfrac{x}{2}\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + C$
4) $\int {\sqrt {{x^2} + {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}} + \dfrac{{{a^2}}}{2}\log \left( {x + \sqrt {{x^2} + {a^2}} } \right) + C$
5) $\int {\sqrt {{x^2} - {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}} - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$