Choose the correct answer in Exercise. 1+x^2 dx is equal to 1. x 2 1+x^2 +1 2 log | (x+ 1+x^2 ) |+C 2. 2 3 (1+x^2)^ 3 2 +C 3. 2 3 x(1+x^2)^ 3 2 +C 4. x^2 2 1+x^2 +1 2 x^2log |x+ 1+x^2 |+C

1. x 2 1+x^2 +1 2 log | (x+ 1+x^2 ) |+C 1+x^2 dx= x^2+1^2 dx= x 2 x^2+1^2 +1^2 2 log |x+ x^2+1^2 |+C = x 2 x^2+1 +1 2 log |x+ x^2+1 |+C [ x^2+a^2 dx= x 2 x^2+a^2 + a^2 2 log |x+ x^2+a^2 | ]+c

Choose the correct answer in Exercise. 1+x^2 dx is equal to 1. x …

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Question

Choose the correct answer in Exercise.
$\int\sqrt{1+\text{x}^2}\text{dx}$ is equal to

$\frac{\text{x}}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{log}\Bigg|\Big(\text{x}+\sqrt{1+\text{x}^2}\Big)\Bigg|+\text{C}$
$\frac{2}{3}(1+\text{x}^2)^{\frac{3}{2}}+\text{C}$
$\frac{2}{3}\text{x}(1+\text{x}^2)^{\frac{3}{2}}+\text{C}$
$\frac{\text{x}^2}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{x}^2\text{log}\Bigg|\text{x}+\sqrt{1+\text{x}^2}\Bigg|+\text{C}$

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Answer

$\frac{\text{x}}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{log}\Bigg|\Big(\text{x}+\sqrt{1+\text{x}^2}\Big)\Bigg|+\text{C}$

$\int\sqrt{1+\text{x}^2}\text{dx}=\int\sqrt{\text{x}^2+1^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+1^2}+\frac{1^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+1^2}\Big|+\text{C}$
$=\frac{\text{x}}{2}\sqrt{\text{x}^2+1}+\frac{1}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+1}\Big|+\text{C}$
$\Bigg[\therefore\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+\text{a}^2}+\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|\Bigg]+\text{c}$

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