How do you integrate #{x(sqrt(1+x^2)} dx# from 0 to 1?

Answer 1

#I=1/3(2sqrt2-1)#.

Let #I=int_0^1 {xsqrt(1+x^2)}dx.#
We subst. #1+x^2=t^2 rArr 2xdx=2tdt, i.e., xdx=tdt#
Further, #x=0 rArr t=1, and, x=1 rArr t=sqrt2#.
#:. I=int_1^(sqrt2) t.tdt = int_1^(sqrt2) t^2dt=[t^3/3]_1^sqrt2#.
#=1/3[sqrt2^3-1]#.
#:. I=1/3(2sqrt2-1)#.
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Answer 2

To integrate ( x\sqrt{1+x^2} ) with respect to ( x ) from 0 to 1, you can use the trigonometric substitution method. Let ( x = \sinh(t) ), where ( \sinh(t) = \frac{e^t - e^{-t}}{2} ). Then, ( dx = \cosh(t) dt ).

Substitute ( x = \sinh(t) ) and ( dx = \cosh(t) dt ) into the integral:

[ \int x\sqrt{1+x^2} , dx = \int \sinh(t) \sqrt{1+\sinh^2(t)} , \cosh(t) , dt ]

[ = \int \sinh(t) \sqrt{\cosh^2(t)} , \cosh(t) , dt ]

[ = \int \sinh(t) \cosh(t)^2 , dt ]

[ = \int \sinh(t) (1 + \sinh^2(t)) , dt ]

[ = \int (\sinh(t) + \sinh^3(t)) , dt ]

[ = \int \sinh(t) , dt + \int \sinh^3(t) , dt ]

[ = \cosh(t) + \frac{1}{3} \cosh^3(t) + C ]

Now, we need to evaluate the integral limits from 0 to 1:

[ \int_0^1 x\sqrt{1+x^2} , dx = \left[ \cosh(t) + \frac{1}{3} \cosh^3(t) \right]_0^1 ]

[ = \left[ \cosh^{-1}(x) + \frac{1}{3} \cosh^3(\cosh^{-1}(x)) \right]_0^1 ]

[ = \left[ \cosh^{-1}(1) + \frac{1}{3} \cosh^3(\cosh^{-1}(1)) \right] - \left[ \cosh^{-1}(0) + \frac{1}{3} \cosh^3(\cosh^{-1}(0)) \right] ]

[ = \left[ \ln(1 + \sqrt{2}) + \frac{1}{3} (1 + \sqrt{2})^3 \right] - \left[ 0 + \frac{1}{3} \cdot 1^3 \right] ]

[ = \ln(1 + \sqrt{2}) + \frac{1}{3} (1 + \sqrt{2})^3 - \frac{1}{3} ]

[ = \ln(1 + \sqrt{2}) + \frac{1}{3} (1 + 3\sqrt{2} + 3 \cdot 2 + \sqrt{2}^3) - \frac{1}{3} ]

[ = \ln(1 + \sqrt{2}) + \frac{1}{3} (1 + 3\sqrt{2} + 3 \cdot 2 + 2\sqrt{2}) - \frac{1}{3} ]

[ = \ln(1 + \sqrt{2}) + \frac{1}{3} (4 + 7\sqrt{2}) - \frac{1}{3} ]

[ = \ln(1 + \sqrt{2}) + \frac{4}{3} + \frac{7}{3}\sqrt{2} - \frac{1}{3} ]

[ = \ln(1 + \sqrt{2}) + \frac{1}{3} + \frac{7}{3}\sqrt{2} ]

Therefore, the value of the integral ( \int_0^1 x\sqrt{1+x^2} , dx ) is ( \ln(1 + \sqrt{2}) + \frac{1}{3} + \frac{7}{3}\sqrt{2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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