Dgdrxrt

Could anyone please give me a hint on how to compute the following integral?

$$int sqrtfracx-2x^7 , mathrm d x$$

I'm not required to use hyperbolic/ inverse trigonometric functions.

Write your integrand in the form $$fracsqrtx-2x^7/2$$ and then substitute $$u=sqrtx$$ so you will get $$2intfracsqrtu^2-2u^6du$$ after this substitute $$u=sqrt2sec(s)$$ to get $$2sqrt2intfracsin^2(s)cos^2(s)4sqrt2ds$$

Write $y(x):=sqrtfrac x-2 x^7$.

Note that $$y'(x)= frac 7-3xx^8 frac 1 y $$

Hence $$frac d dx x^n y=n x^n-1 y + x^n-8 frac 7-3x y.$$

Do the ansatz $$F(x)=sum_n=0^k a_nx^ny ~~~~text and ~~~~F'(x)=y(x). $$

We get
$$ysum_n=1^ka_n nx^n-1 +frac 1 ysum_n=0^k a_n x^n-8 (7-3x)=y $$

Thus $$sum_n=0^k a_n x^n-8 (7-3x)=y^2(1-sum_n=1^ka_n nx^n-1).$$

Now insert $y^2$
and get
$$(7-3x)sum_n=0^k a_n x^n-1 = (x-2 )(1-sum_n=1^ka_n nx^n-1) $$
or equivalently
$$2-x+(7-3x)sum_n=0^k a_n x^n-1+ (x-2)sum_n=1^ka_n nx^n-1=0. $$

We solve this recursively.

The lowest order is $x^-1$. There we have $7a_0 x^-1=0$, so $a_0=0$.

In constant order: $2+7a_1-3a_0-2a_1=0$, so $a_1=-frac 2 5$

In order $x$: $(-1+7a_2 -3 a_1 +a_1-4 a_2)x=0$ , so $a_2= frac 1 3 (1+2
a_1)=frac 1 15$

in order $x^2$: $(7a_3-3a_2+2a_2-6 a_3)x^2=0$, so $a_3=a_2=frac 1 15$.

in order $x^3$: $(7a_4-3a_3+3a_3 - 8a_4)x^4=0$, so $a_4=0$

in order $x^n$ for $n>3$: $(7a_n+1 - 3 a_n +n a_n - 2(n+1) a_n+1)x^n$, so $a_n+1=frac n-37-2(n+1) a_n=0$.

In conclusion it follows that
$$int y(x)= const + F(x)= const+ frac 1 15 (-6x+ x^2+x^3) y $$

With $y:=dfrac1x$,

$$intsqrtfracx-2x^7dx=-intsqrtleft(frac1y-2right)y^7,fracdyy^2=-int ysqrt1-2y,dy.$$

Then by parts,

$$-int ysqrt1-2y,dy=frac13y(1-2y)^3/2-frac13int(1-2y)^3/2dy=frac13y(1-2y)^3/2+frac115(1-2y)^5/2$$

$$=frac13xleft(1-frac2xright)^3/2+frac115left(1-frac2xright)^5/2.$$

Try the substitution
$$
u=fracx-2x
$$

or equivalently

$$
x=frac21-u
$$