Let x²=t
Differentiating w.r.t x, we get
2x dx=dt
?∫ 2x² e^x2 dx=∫ te^t dt
By using formula, ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) ,we get
∫ t e^t dt=t∫ e^t dt-∫ (t)’∫ e^t dt
=te^t-∫ e^t dt
=te^t-e^t=e^t (t-1)
Replacing t with x², we get
∫ 2x³ e^x2 dx=e^x2 (x²-1)+C

Let \(I=\int_0^?(1-sin?3x)dx\)
F(x)=∫ 1-sin?3x dx
=x+\(\frac{cos?3x}{3}\)
Applying limits by using fundamental orem of calculus, we get
I=F(?)-F(0)
=\(?+\frac{cos?3?}{3}-0-\frac{cos?0}{3}\)
=\(?-\frac{1}{3}-\frac{1}{3}=?-\frac{2}{3}=\frac{3?-2}{3}\).

It is a proper rational function. Therefore,
\(\frac{1}{(x+1)(x+2)} = \frac{A}{(x+1)} + \frac{B}{(x+2)}\)
Where real numbers are determined, 1 = A(x+2) + B(x+1), Equating coefficients of x and constant term, we get A+B = 0 and 2A+B = 1. Solving it we get A=1, and B=-1.
Thus, it simplifies to, \(\frac{1}{(x+1)} + \frac{-1}{(x+2)} = \int \frac{dx}{(x+1)} – \int \frac{dx}{(x+2)}\).
= log|x+1| – log|x+2| + C
= \(Log \left|\frac{x+1}{x+2}\right|+ C\).

If we solve L.H.S. (Left Hand Side) of equation, we get following value.
(x+2) (x+4) = x² + 4x + 2x + 8 = x² + 6x + 8.
This value is same as R.H.S. (Right Hand Side).
So, all values of x satisfy equality.

We know that, sin?3x=3 sin?x-4 sin³?x
?sin³x=\(\frac{3 sin?x-sin?3x}{4}\)
\(\int 2 \,sin^3?x+1 \,dx=\int \frac{(3 sin?x-sin?3x)}{2} dx+\int dx\)
=\(\frac{3}{2} \int sin?x dx-\frac{1}{2} \int sin?3x dx+\int dx\)
=-\(\frac{3}{2} cos?x+\frac{cos?3x}{6}+x+C\)

\(\int 3e^x+\frac{2(log?x^2)}{3x} dx=3\int e^x dx+\frac{2}{3} \int \frac{log?x}{x}\)
Let log?x=t
Differentiating w.r.t x, we get
\(\frac{1}{x} dx=dt\)
?\(\int \frac{log?x}{x}=\int \,t \,dt=\frac{t^2}{2}\)
\(\int e^x dx=e^x\)
Replacing t with log?x, we get
\(\int 3e^x+\frac{2(log?x^2)}{3x} dx=3e^x+\frac{1}{3} (log?x)^2+C\)

To find \(\int \,3 \,cos?x+\frac{1}{x^2} dx\)
\(\int \,3 \,cos?x+\frac{1}{x^2} dx=3 \int cos?x \,dx+\int \frac{1}{x^2} \,dx\)
\(\int \,3 \,cos?x+\frac{1}{x^2} dx=3 \,sin?x+\int x^{-2} \,dx\)
\(\int \,3 \,cos?x+\frac{1}{x^2} dx=3 \,sin?x+\frac{x^{-2+1}}{-2+1}\)
\(\int \,3 \,cos?x+\frac{1}{x^2} dx=3 \,sin?x-\frac{1}{x}+C\)

Let \(I=\int_0^{\frac{?}{2}} \,5 \,sin?x \,dx\)
F(x)=\(\int5 \,sin?x \,dx=-5 \,cos?x\)
Applying limits by using fundamental orem of calculus, we get
I=F(\(\frac{?}{2}\))-F(0)
&re4;\(\int_0^{\frac{?}{2}} \,5 \,sin?x \,dx=-5[cos?\frac{?}{2}-cos?0]\)
=-5[0-1]=5

\(\int \frac{dx}{\sqrt{5-x^2}}=\int \frac{dx}{\sqrt{(?5)^2-x^2}}\)
By using formula \(\int \frac{dx}{\sqrt{a^2-x^2}}=sin^{-1}?\frac{x}{a}+C\)
?\(\int \frac{dx}{\sqrt{x^2-5}}=sin^{-1}?\frac{x}{\sqrt{5}}+C\)

\(\int_3^7\)2f(x)-g(x)dx = 2 \(\int_3^7\)f(x) – \(\int_3^7\)g(x)dx
= 2(4) ? 2
= 6