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: $\displaystyle \int\left(\frac{1}{(x-1)^{2}}+\frac{x}{x^{2}+1}\right)dx$

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

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

3. Hello, simfonija!

Here's my approach to #1 . . .

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

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

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

Let $\displaystyle 2t+1 \,=\,2\sqrt{3}\sec\theta\quad\Rightarrow\quad t \,=\,\frac{1}{2}(2\sqrt{3}\sec\theta - 1)\quad\Rightarrow\quad dt$$\displaystyle \,= \,\sqrt{3}\sec\theta\tan\theta\,d\theta and the radical is: .\displaystyle \sqrt{(2t+1)^2-12}\:=\:\sqrt{(2\sqrt{3}\sec\theta)^2 - 12} . . . \displaystyle =\:\sqrt{12\sec^2\theta - 12} \:=\:\sqrt{12(\sec^2\theta - 1)} \displaystyle \:=\:\sqrt{12\tan^2\theta} \:=\:2\sqrt{3}\tan\theta Substitute: .\displaystyle 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) \;= \displaystyle \frac{3}{4}\int\sec\theta(2\sqrt{3}\sec\theta - 1)\,d\theta . . \displaystyle = \;\frac{3}{4}\int\left(2\sqrt{3}\sec^2\theta - \sec\theta)\,d\theta \:= \displaystyle \:\frac{3}{4}\left[2\sqrt{3}\tan\theta - \ln\left|\sec\theta + \tan\theta\right|\right] + C Back-substitute: .\displaystyle \sec\theta \:=\:\frac{2t + 1}{2\sqrt{3}}\quad\Rightarrow\quad \tan\theta \:=\:\frac{\sqrt{4t^2 + 4t - 11}}{2\sqrt{3}} We have: .\displaystyle \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 . . \displaystyle = \;\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, \displaystyle \int e^{\sqrt{x}} dx Express as, \displaystyle 2\int \frac{\sqrt{x}}{2\sqrt{x}}e^{ \sqrt{x} } dx Let, \displaystyle u=\sqrt{x} then, \displaystyle \frac{du}{dx}=\frac{1}{2\sqrt{x}} Thus, you have, \displaystyle 2\int \sqrt{x} e^{\sqrt{x}}\cdot \frac{1}{2\sqrt{x}} dx Becomes, \displaystyle 2\int u e^u \frac{du}{dx}dx=2\int ue^udu Ignore the 2 and use integration by parts, \displaystyle v=u then, \displaystyle v'=1 \displaystyle w'=e^u then, \displaystyle w=e^u Thus, by parts, \displaystyle ue^u-\int e^u du Thus, \displaystyle ue^u-e^u+C Multiply by two, \displaystyle 2ue^u-2e^u+C Substitute, \displaystyle 2\sqrt{x} e^{\sqrt{x}} -2e^{\sqrt{x}}+C 5. Hello, simfonija! Galactus had the best idea for #3 . . . \displaystyle 3)\;\;\int e^{\sqrt{x}}\,dx Let \displaystyle \sqrt{x} = y\quad\Rightarrow\quad x = y^2\quad\Rightarrow\quad dx = 2y\,dy Substitute: .\displaystyle \int e^y(2y\,dy) \;= \;2\int y\,e^y\,dy Integrate by parts: . . Let: . \displaystyle u = y\qquad dv = e^y\,dy . . Then: \displaystyle du = dy\quad v = e^y And we have: .\displaystyle 2\left[y\!\cdot\! e^y - \int e^u\,du\right] \;=\;2\left(y\!\cdot\! e^y - e^y\right) + C \displaystyle = \;2e^y(y - 1) + C Back-substitute: .\displaystyle \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: \displaystyle \frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{Cx+D}{x^{2}+1} \displaystyle (x^{2}+1)(x-1)A+B(x^{2}+1)+(Cx+D)(x-1)^{2}$$\displaystyle =x^{3}-x^{2}+x+1$

This leads to the following system:

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

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

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

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

For the first half:

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

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

For the 2nd half:

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

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

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

So, put 'er all together:

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

8. ## simfonija

Thanks again, great help.