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wrote...
5 months ago
 Im stuck at integrating algebraic substitution Can you solve this pls. By using algebraic substitution.Integral of x^2/(x^2+1)^3/2 Read 143 times 1 Reply
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wrote...
Educator
5 months ago
 Our function is:$$\int \frac{x^2}{\left(x^2+1\right)^{\frac{3}{2}}}dx$$I looked at this for 20 minutes, and realized that algebraic substitution is NOT the right way to do it.Instead, you have to perform trigonometric substitution:$$x=\tan \left(u\right)$$Solve for $$u$$, then find derivative with respect to x:$$\tan ^{-1}x=u$$$$\frac{du}{dx}=\sec ^2\left(u\right)$$Solve for dx:$$du=\sec ^2\left(u\right)\cdot dx$$Now your expression is:$$\int \frac{\tan ^2\left(u\right)}{\left(\tan ^2u+1\right)^{\frac{3}{2}}}\sec ^2\left(u\right)du$$At the bottom:Simplify using $$\tan ^2(u)+1=\sec ^2(u)$$We get:$$\int \frac{\tan ^2\left(u\right)}{\left[\sec ^2\left(u\right)\right]^{\frac{3}{2}}}\sec ^2\left(u\right)du$$$$\int \frac{\tan ^2\left(u\right)}{\sec ^3\left(u\right)}\sec ^2\left(u\right)du$$$$\int \frac{\tan ^2\left(u\right)}{\sec \left(u\right)}du$$Same as:$$\int \cos \left(u\right)\tan ^2\left(u\right)du$$...Please let me know if you're following
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