Find $\int\sqrt{t^{3/2}+1}\,dt$

Question

Integrate $\int\sqrt{t^{3/2}+1}\,dt$

I tried all possible substituion but its not worked. Wolfram alpha also gives something different answer

Can someone give me just a hint please?

Is there some reason to expect anything else than the hypergeometric function WA provides, obtained by expanding the square root and integrating each power $t^{3n/2}$? — Did, Oct 10, 2017 at 8:45
Maybe trying something like t = (tanx)^4/3 or something should help — Sagar Chand, Oct 10, 2017 at 8:47
Try t = x^2 you have a: $\int x \sqrt{1+x^3} \, dx$ and integrating by part. — Mariusz Iwaniuk, Oct 10, 2017 at 11:35
Hint: it is not an elementary function. Use series expansion to get the answer in terms of a hypergeometric function — GEdgar, Oct 10, 2017 at 14:28
Suppose $(x^{3/2} +1)^{0.5}= Sin a$, we have: $$ (Sin^2 a - 1 )^{2/3}= (Cos a)^{4/3}$$ $$ A_1 = \int Cos^{5/3}a da $$ Now we can use this formula to find $A_1$: $$\int Cos^n x dx = \frac{Sin ^{n-1} x Sin x}{n} + \ frac {n-1}{n}\int Cos ^{n-2} x dx $$ I do not think $(x^{3/2} +1)^{0.5}$ is integrable by conventional ways. — sirous, Oct 14, 2017 at 7:42

Harry Peter · Accepted Answer · 2017-10-16 12:34:46Z

When $|t|\leq1$ ,

$\int\sqrt{t^\frac{3}{2}+1}~dt$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n}{2}}{4^n(n!)^2(1-2n)}dt$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n+2}{2}}{4^n(n!)^2(1-2n)\dfrac{3n+2}{2}}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n+2}{2}}{2^{2n-1}(n!)^2(1-2n)(3n+2)}+C$

When $|t|\geq1$ ,

$\int\sqrt{t^\frac{3}{2}+1}~dt$

$=\int t^\frac{3}{4}\sqrt{1+\dfrac{1}{t^\frac{3}{2}}}~dt$

$=\int t^\frac{3}{4}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)t^\frac{3n}{2}}dt$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3-6n}{4}}{4^n(n!)^2(1-2n)}dt$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{7-6n}{4}}{4^n(n!)^2(1-2n)\dfrac{7-6n}{4}}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^{n-1}(n!)^2(2n-1)(6n-7)t^\frac{6n-7}{4}}+C$

1 Answer 1