Let \(I=\int_{-1}^23x+x^2-2 \,dx\)
F(x)=\(\int 3x+x^2-2 \,dx\)
=\(\frac{3x^2}{2}+\frac{x^3}{3}-2x\)
Applying limits, we get
I=F(2)-F(-1)
I=\(\left(\frac{(3�2^3)}{2}+\frac{2^3}{3}-2(2)\right)-\left(\frac{3 (-1)^2}{2}+\frac{(-1)^3}{3}-2(-1)\right)\)
I=\(6+\frac{8}{3}-4-\frac{3}{2}+\frac{1}{3}-2=\frac{3}{2}\).

I=\(\int_0^1 \frac{5 \,sin?(tan^{-1)}x}{1+x^2} \,dx\)
Let tan^-1?x=t
Differentiating w.r.t x, we get
\(\frac{1}{1+x^2} \,dx=dt\)
The new limits
When x=0, t=tan^-1?0=0
When x=1, t=tan^-1)1=?/4
?\(\int_0^1 \frac{5 \,sin?(tan^{-1}?x)}{1+x^2} \,dx=\int_0^{?/4} \,5 \,sin?t \,dt\)
=\(5[-cos?t]_0^{?/4}=-5[cos?t]_0^{?/4}\)
\(=-5(cos? \frac{?}{4}-cos?0)=-5(\frac{1}{\sqrt{2}-1})=5-\frac{5}{\sqrt{2}}\)

\(\int_3^6\)9 e^x dx = 9(e^x)⁶₃ dx
= 9(e⁶ ? e³)

Let \(I=\int_{?/4}^{?/2}7 \,cos?x \,dx\)
F(x)=∫ 7 cos?x dx
=7(sin?x)
Applying limits by using second fundamental orem of calculus, we get
\(I=F(\frac{?}{2})-F(\frac{?}{4})=7(sin\frac{?}{2}-sin? \frac{?}{4})=7(1-\frac{1}{\sqrt{2}})\)

We know that \(\int \frac{dx}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}\)
By simplifying it we get, \(\int \frac{dx}{x(x^2+1)}=\frac{(A+B) x^2+Cx+A}{x(x^2+1)}\)
Now equating coefficients we get A = 0, B = 0, C=1.
\(\int \frac{dx}{x(x^2+1)} = \int \frac{dx}{x} + \int \frac{-xdx}{(x^2+1)}\)
Therefore after integrating we get \(log|x| – \frac{1}{2} log(x^2+1)\) + C.

\(\int_8^2\)2f(x)dx = -2 \(\int_2^8\)f(x)dx
= – 2(-3)
= 6

Let x⁵+9=t
Differentiating w.r.t x, we get
5x⁴ dx=dt
\(\int \frac{5x^4}{\sqrt{x^5+9}} dx=\int \frac{dt}{\sqrt{t}}\)
=\(\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2\sqrt{t}\)
Replacing t with x⁵+9, we get
\(\int \frac{5x^4}{\sqrt{x^5+9}} dx=2\sqrt{x^5+9}\).

To find \(\int \,3e^x+\frac{2}{x}+x^3 \,dx\)
\(\int \,3e^x+\frac{2}{x}+x^3 dx=3\int \,e^x \,dx+2\int \frac{1}{x} \,dx+\int x^3 \,dx\)
\(\int \,e^x \,dx=e^x\)
\(\int \frac{1}{x} dx=log?x\)
&re4;\(\int 3e^x+\frac{2}{x}+x^3 \,dx=3e^x+2 \,log?x+\frac{x^4}{4}+c\)

\(\int \frac{dx}{x^2+4}=\int \frac{dx}{x^2+2^2}\)
Using formula \(\int \frac{dx}{a^2+x^2}=\frac{1}{a} tan^{-1}?\frac{x}{a}+C\)
?\(\int \frac{dx}{x^2+2^2}=(\frac{1}{2} tan^{-1}?\frac{x}{2})+C\)

Let cos?x=t
Differentiating w.r.t x, we get
-sin?x dx=dt
sin?x dx=-dt
?∫ sin?x log?(cos?x) dx=∫ -log?t dt
Using ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) , we get
∫ -log?t dt=-log?t ∫ 1 dt-∫ (-log?t)’ ∫ 1 dt
=-t log?t+∫ dt
=-t log?t+t
=t(log?t-1)
Replacing t with cos?x, we get
∫ sin?x log?(cos?x) dx=cos?x (log?(cos?x)-1)+C