Question:
\[ \int \sqrt{x^2+3 x} d x= \]
\[ \int \sqrt{x^2+3 x} d x= \]
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Complete the square under the root and use the standard form for the integral of the square root of u squared minus a squared.
Updated On: Nov 13, 2026
\(\sqrt{x^{2}+3x}+\mathrm{log}\sqrt{x^{2}+3x}+c\), where c is the constant of integration.
\(\frac{2x+3}{4}\sqrt{x^{2}+3x}-\frac{9}{8}\mathrm{log}(x+\sqrt{x^{2}+3x})+c\), where c is the constant of integration.
\(x\sqrt{x^{2}+3x}+\mathrm{log}(x+\sqrt{x^{2}+3x})+\mathrm{c}\), where c is the constant of integration.
\(x+3\sqrt{x^{2}+3x}+\frac{3}{2}\mathrm{log}(x+\sqrt{x^{2}+3x})+\mathrm{c}\), where c is the constant of integration.
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Write \(x^2+3x=\left(x+\dfrac32\right)^2-\dfrac94\). Then use \(\int\sqrt{u^2-a^2}\,du=\dfrac u2\sqrt{u^2-a^2}-\dfrac{a^2}{2}\log\left|u+\sqrt{u^2-a^2}\right|\).
Step 2: Substitute
Here \(u=x+\dfrac32\) and \(a=\dfrac32\), so \(\dfrac u2=\dfrac{2x+3}{4}\) and \(\dfrac{a^2}{2}=\dfrac98\).
Step 3: Calculation
\(\int\sqrt{x^2+3x}\,dx=\dfrac{2x+3}{4}\sqrt{x^2+3x}-\dfrac98\log\left|x+\dfrac32+\sqrt{x^2+3x}\right|+c\)
Step 4: Compare the options
Only option (B) has the factors \(\dfrac{2x+3}{4}\) and \(\dfrac98\), so it is the intended answer.
Final Answer:
The integral has the factor (2x+3)/4 times the root minus 9/8 times a logarithm. \[ \boxed{\text{(B) }\dfrac{2x+3}{4}\sqrt{x^{2}+3x}-\dfrac{9}{8}\log\left(x+\sqrt{x^{2}+3x}\right)+c} \]
Write \(x^2+3x=\left(x+\dfrac32\right)^2-\dfrac94\). Then use \(\int\sqrt{u^2-a^2}\,du=\dfrac u2\sqrt{u^2-a^2}-\dfrac{a^2}{2}\log\left|u+\sqrt{u^2-a^2}\right|\).
Step 2: Substitute
Here \(u=x+\dfrac32\) and \(a=\dfrac32\), so \(\dfrac u2=\dfrac{2x+3}{4}\) and \(\dfrac{a^2}{2}=\dfrac98\).
Step 3: Calculation
\(\int\sqrt{x^2+3x}\,dx=\dfrac{2x+3}{4}\sqrt{x^2+3x}-\dfrac98\log\left|x+\dfrac32+\sqrt{x^2+3x}\right|+c\)
Step 4: Compare the options
Only option (B) has the factors \(\dfrac{2x+3}{4}\) and \(\dfrac98\), so it is the intended answer.
Final Answer:
The integral has the factor (2x+3)/4 times the root minus 9/8 times a logarithm. \[ \boxed{\text{(B) }\dfrac{2x+3}{4}\sqrt{x^{2}+3x}-\dfrac{9}{8}\log\left(x+\sqrt{x^{2}+3x}\right)+c} \]
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