I'm trying to find the indefinite integrals for the two equations below. What methods do I use to solve? (-1 < x < 1) Thanks
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The first one can be solved by integration by parts, let $\displaystyle u = arctan x, dv = 2x^3 + x$ Second one can be done easily by a substitution of $\displaystyle u = 1 - x^2$
Originally Posted by jennybee I'm trying to find the indefinite integrals for the two equations below. What methods do I use to solve? (-1 < x < 1) Thanks For the first one use integration by parts $\displaystyle u=\tan^{-1}(x) \quad dv=2x^3+x$ For the 2nd one use a u du substitution $\displaystyle u=1-x^2$
Thanks to both of you. Loving the Sephiroth avatar
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