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How do i start off with integrating the below function? i tried applying trig substitution and U substitution. how do i go about solving this function? should i split them up further into 2 separate functions ? need some help in this as i can't seem to figure out how to continue on with it

$$\int\frac{x^{3}}{({x^{2}-1})^{0.5}}dx$$

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3 Answers 3

Rewrite the integrand as $\frac{x(x^2-1)}{(x^2-1)^{1/2}}+\frac{x}{(x^2-1)}$, which is $x(x^2-1)^{1/2}+\frac{x}{(x^2-1)^{1/2}}$. Then $u=x^2-1$ finishes things.

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Note that chosing $u(x) = x^2$ and $v(x) = (x^2-1)^{\frac{1}{2}}$, we have

$$\int u(x)v'(x)\, dx = \int\frac{x^{3}}{({x^{2}-1})^{0.5}}dx\quad\text{(check this!)} $$

where $v'(x)$ denotes the first derivative by $x$. Thus we can use integration by parts, i.e. the formula

$$\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx$$

The rest are relatively easy calculations

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$$I=\int\frac{x^3}{\sqrt{x^2-1}}dx=\int\frac{x^2}{\sqrt{x^2-1}}x\ dx$$

Setting $\displaystyle\sqrt{x^2-1}=y,\frac{2x}{2\sqrt{x^2-1}}dx=dy$ and $\displaystyle x^2=y^2+1$

$$I=\int(y^2+1)dy=\cdots $$


Alternatively,

Using Trigonometric substitution (1,2), $\displaystyle x=\sec y\implies dx=\sec y\tan y\ dy$

$$\int\frac{x^3}{\sqrt{x^2-1}}dx=\int\frac{\sec^3y}{\tan y}\sec y\tan y\ dy=\int\sec^4y\ dy$$

$$=\int(\tan^2y+1)\sec^2y\ dy$$

Now set $\displaystyle\tan y =u$

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