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6

This is a cute $u$-substitution problem. The crux of the problem is dividing out by the right factor of $x$. Notice that if we factor out one copy of $x$ we obtain $$\int \frac{1}{x\cdot x^{\frac{16}{25}} + x^{\frac{9}{25}} }dx \;\; =\;\; \int \frac{1}{x \left (x^{\frac{16}{25}} + x^{\frac{-16}{25}} \right )}dx.$$ Now let $u = x^{\frac{16}{25}}$, and ...

6

The function $\frac{e^x}x$ has no elementary primitive. So, no, it is not a standard integral.

3

Yes, there is a way without partial fraction. This might also be in your interest. Start off with the substitution: $$x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$$ This substitution produces a nice cancelation in the denominator, since: $$(1+t)^3+(1-t)^3=1+3t +3t^2+t^3 +1-3t+3t^2 -t^3=2(1+3t^2)$$ $$\int \frac{1}{1+x^3}dx=-\int\frac{1}{\frac{(1+t)^... 3 That notation is used in the classic textbook Elementary Differential Equations by William E. Boyce and Richard C. DiPrima, at least in the third edition (1976), which is the one I have. Quoting from p. 11 (beginning of Chapter 2): The simplest type of first order differential equation occurs when f depends only on x. ... 3 In general, that integral diverges. That will be the case if, say, x_0=0, f(x)=x-1 and g(x)=x. So, yes, some extra hypothesis is required here. Since (assuming that the definition of h makes sense) h'(x)=\dfrac{f(x)}{g(x)}, and since this quotient is smaller than 0 on [x_0,x_1) and greater than 0 on (x_1,\infty), you are right. 2 Let$$x=y^{25}\implies dx=25\,y^{24}\,dy\int\frac{dx}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}=25\int\frac{ y^{15}}{y^{32}+1}\,dy=\frac {25}{16}\int\frac{16\, y^{15}}{y^{32}+1}\,dy=\frac {25}{16}\int\frac{ \left(y^{16}\right)'}{\left(y^{16}\right)^2+1}$$2 You can apply any substitution to any integral; you may have seen this particular one recommended for integrals of the form you mention, but that doesn't mean you can't use it elsewhere as well. Of course, to use a substitution, you need to satisfy the hypotheses of whatever theorem you have that allows substitutions, but in practice, we often proceed by ... 1 The integral \int \frac{e^x}{x}dx cannot be expressed with a finite number of elementary functions. A closed form requires a special function named Ei(x): http://mathworld.wolfram.com/ExponentialIntegral.html$$\int \frac{e^x}{x}dx=\text{Ei}(x)+\text{constant}.\int \frac{e^x}{x^2}dx=\text{Ei}(x)-\frac{e^x}{x}+\text{constant}. If you don't need ...

1

Two methods to solve integrals with the form $e^x x^{-n} dx$: Either use the power series of the exponential function, then integrate, then sum back Do $n$ integration by parts. As you may have noticed, you reduce the order of the exponent of $x$ by one at each integration by parts. Also please note that $\int e^x x^{-1}dx$ is called the exponential ...

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