I need solution step by step for these integrals
I have Maple but don't have options show me step by step
Could you help
Thanks

1. int(3*t/(4*t^2+4*t-11)^1/2,t)
2. int((x^3-x^2+x+1)/(x^2+1)*(x-1)^2,x);
3. int(exp(((x)^1/2)),x);

2. Here's a link to the first one. Was that you?.

#2 use partial fractions: $\int\left(\frac{1}{(x-1)^{2}}+\frac{x}{x^{2}+1}\right)dx$

For #3, try the subsitution $u=\sqrt{x}$

$u^{2}=x\;\; 2udu=dx$

3. Hello, simfonija!

Here's my approach to #1 . . .

$1)\;\;\int \frac{3t}{\sqrt{4t^2 + 4t - 11}}\,dt$

Complete the square: . $4t^2 + 4t - 11\;=\;4t^2 + 4t + 1 - 11\;=\;(2t + 1)^2 - 12$

So we have: . $3\int\frac{t}{\sqrt{(2t+1)^2 - 12}}\,dt$

Let $2t+1 \,=\,2\sqrt{3}\sec\theta\quad\Rightarrow\quad t \,=\,\frac{1}{2}(2\sqrt{3}\sec\theta - 1)\quad\Rightarrow\quad dt$ $\,= \,\sqrt{3}\sec\theta\tan\theta\,d\theta$

and the radical is: . $\sqrt{(2t+1)^2-12}\:=\:\sqrt{(2\sqrt{3}\sec\theta)^2 - 12}$

. . . $=\:\sqrt{12\sec^2\theta - 12} \:=\:\sqrt{12(\sec^2\theta - 1)}$ $\:=\:\sqrt{12\tan^2\theta} \:=\:2\sqrt{3}\tan\theta$

Substitute: . $3\int\frac{\frac{1}{2}(2\sqrt{3}\sec\theta - 1}{2\sqrt{3}\tan\theta}\,(\sqrt{3}\sec\theta\tan \theta\,d \theta) \;=$ $\frac{3}{4}\int\sec\theta(2\sqrt{3}\sec\theta - 1)\,d\theta$

. . $= \;\frac{3}{4}\int\left(2\sqrt{3}\sec^2\theta - \sec\theta)\,d\theta \:=$ $\:\frac{3}{4}\left[2\sqrt{3}\tan\theta - \ln\left|\sec\theta + \tan\theta\right|\right] + C$

Back-substitute: . $\sec\theta \:=\:\frac{2t + 1}{2\sqrt{3}}\quad\Rightarrow\quad \tan\theta \:=\:\frac{\sqrt{4t^2 + 4t - 11}}{2\sqrt{3}}$

We have: . $\frac{3}{4}\bigg[2\sqrt{3}\cdot\frac{\sqrt{4t^2+4t-11}}{2\sqrt{3}} - \ln\left|\frac{2t+1}{2\sqrt{3}} + \frac{\sqrt{4t^2+4t-11}}{2\sqrt{3}}\right|\,\bigg] + C
$

. . $= \;\frac{3}{4}\bigg[\sqrt{4t^2+4t-11} - \ln\left|2t + 1 + \sqrt{4t^2 + 4t - 11\right|\,\bigg] + C$

4. Originally Posted by simfonija
3. int(exp(((x)^1/2)),x);
You have,
$\int e^{\sqrt{x}} dx$
Express as,
$2\int \frac{\sqrt{x}}{2\sqrt{x}}e^{ \sqrt{x} } dx$
Let, $u=\sqrt{x}$ then, $\frac{du}{dx}=\frac{1}{2\sqrt{x}}$
Thus, you have,
$2\int \sqrt{x} e^{\sqrt{x}}\cdot \frac{1}{2\sqrt{x}} dx$
Becomes,
$2\int u e^u \frac{du}{dx}dx=2\int ue^udu$
Ignore the 2 and use integration by parts,
$v=u$ then, $v'=1$
$w'=e^u$ then, $w=e^u$
Thus, by parts,
$ue^u-\int e^u du$
Thus,
$ue^u-e^u+C$
Multiply by two,
$2ue^u-2e^u+C$
Substitute,
$2\sqrt{x} e^{\sqrt{x}} -2e^{\sqrt{x}}+C$

5. Hello, simfonija!

Galactus had the best idea for #3 . . .

$3)\;\;\int e^{\sqrt{x}}\,dx$

Let $\sqrt{x} = y\quad\Rightarrow\quad x = y^2\quad\Rightarrow\quad dx = 2y\,dy$

Substitute: . $\int e^y(2y\,dy) \;= \;2\int y\,e^y\,dy$

Integrate by parts:

. . Let: . $u = y\qquad dv = e^y\,dy$

. . Then: $du = dy\quad v = e^y$

And we have: . $2\left[y\!\cdot\! e^y - \int e^u\,du\right] \;=\;2\left(y\!\cdot\! e^y - e^y\right) + C$ $= \;2e^y(y - 1) + C$

Back-substitute: . $\boxed{2e^{\sqrt{x}}(\sqrt{x} - 1) + C}$

6. Thank you so much to all, can somebody help with the last one, I get lost in coificients, cant find A,B... I have just 15 min to submit my assignment
Thanks

7. #2:

Do the partial fraction thing:

$\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{Cx+D}{x^{2}+1}$

$(x^{2}+1)(x-1)A+B(x^{2}+1)+(Cx+D)(x-1)^{2}$ $=x^{3}-x^{2}+x+1$

This leads to the following system:

$A+C=1;\\-A+B-2C+D=$ $1;\\A+C-2D+1;\\-A+B+D=1$

The solutions are $A=0;\;\ B+1;\;\ C=1;\;\ D=0$

Break it into 2 integrals after you do the partial fractions.

$\int\frac{1}{(x-1)^{2}}dx+\int\frac{x}{x^{2}+1}dx$

For the first half:

$u=x-1;\;\ u+1=x;\;\ du=dx$

$\int\frac{1}{u^{2}}du=\frac{-1}{u}=\frac{-1}{x-1}$

For the 2nd half:

$\int\frac{x}{x^{2}+1}dx$

$u=x^{2}+1;\;\ du=2xdx;\;\frac{du}{2}=xdx$

$\frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}ln(u)=\fra c{1}{2}ln(x^{2}+1)$

So, put 'er all together:

$\frac{-1}{x-1}+\frac{1}{2}ln(x^{2}+1)$

8. simfonija

Thanks again, great help.