[See Steps] Determine the following integrals. ∫ (6 x^2-6 x)/((1-3 x^2+2 x^3)^3/4) d x ∫[(x)/(3 x^2+5)-fracx^2(x^3+1)^3] d x ∫((2)/(t^2) √1/t+9)
Question: Determine the following integrals.
- \[\int \frac{6 x^{2}-6 x}{\left(1-3 x^{2}+2 x^{3}\right)^{\frac{3}{4}}} d x\]
- \[\int\left[\frac{x}{3 x^{2}+5}-\frac{x^{2}}{\left(x^{3}+1\right)^{3}}\right] d x\]
- \[\int\left(\frac{2}{t^{2}} \sqrt{\frac{1}{t}+9}\right) d t\]
- \[\int \frac{2 x}{\left(x^{2}+1\right) \ln \left(x^{2}+1\right)} d x\]
- \[\int_{-1}^{1}\left(e^{x}-e^{-x}\right)^{2} d x\]
- \[\int_{3}^{4} \frac{2 x^{4}-6 x^{3}+x-2}{x-2} d x\]
- \[\int_{\sqrt{2}}^{\sqrt{7}} 3\left(x^{2}+2\right)^{-\frac{1}{2}} x e^{\sqrt{x^{2}+2}} d x\]
- \[\int_{-2}^{1} 8|x| d x\]
where R, j and \(k\) are constants.
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