Q.$\dfrac{x}{\sqrt{1+x^{2}}}$

Substitute $t=1+x^2$, so $dt=2x\,dx$ — the numerator is exactly $dt$.

Let $t=1+x^2$, then $dt=2x\,dx$.

$\int \frac{2x}{1+x^2}\,dx = \int \frac{dt}{t} = \ln|t|+C = \ln(1+x^2)+C$

(the $+C$ can be dropped from absolute value since $1+x^2>0$ always)

Substituting $t=\sqrt x$ turns the integral into $2\int te^t\,dt$, which by parts gives $2e^{\sqrt x}(\sqrt x-1)+c$.

Let $t=\sqrt{x}$, so $x=t^2$ and $dx=2t\,dt$.

$\displaystyle\int e^{\sqrt{x}}dx=\int e^t\cdot2t\,dt=2\int t\,e^t\,dt$

Using integration by parts with $u=t$, $dv=e^t dt$ (so $du=dt$, $v=e^t$):

$2\int t\,e^t\,dt=2\left[t\,e^t-\int e^t\,dt\right]=2\left[t\,e^t-e^t\right]+c=2e^t(t-1)+c$

Substituting back $t=\sqrt{x}$: $2e^{\sqrt{x}}(\sqrt{x}-1)+c$.

Substituting $u=1/x$ converts the integral into a standard exponential integral.

Let $u=\dfrac{1}{x}$, so $du=-\dfrac{1}{x^2}dx$, i.e. $\dfrac{dx}{x^2}=-du$.

Then $\displaystyle\int\dfrac{3^{1/x}}{x^2}dx=\int 3^u(-du)=-\int 3^u\,du=-\dfrac{3^u}{\log3}+c=-\dfrac{3^{1/x}}{\log3}+c$.

Comparing with $k\left(3^{1/x}\right)+c$, we get $k=-\dfrac{1}{\log3}$.

Use the substitution $u=\log x$; the $\dfrac{1}{x}\,dx$ becomes $du$, leaving a standard power integral. The answer is $\dfrac{1}{2}(\log x)^2+c$, option (b).

Substitution. Let

$u=\log x \implies du=\frac{1}{x}\,dx.$

Rewrite the integral.

$\int \frac{\log x}{x}\,dx=\int u\,du.$

Integrate.

$\int u\,du=\frac{u^2}{2}+c=\frac{1}{2}(\log x)^2+c.$

The other options do not arise from any valid integration of $\dfrac{\log x}{x}$.

The integral equals $\sqrt{\tan x}+C$.

Let $u=\tan x$, so $du=\sec^2x\,dx$, i.e. $dx=\cos^2x\,du$.

Also $\sin2x=\dfrac{2\tan x}{1+\tan^2x}=\dfrac{2u}{1+u^2}$ and $\cos^2x=\dfrac{1}{1+u^2}$.

So the integrand becomes

$\dfrac{\sqrt u}{\frac{2u}{1+u^2}}\cdot\dfrac{1}{1+u^2}\,du=\dfrac{\sqrt u(1+u^2)}{2u(1+u^2)}\,du=\dfrac{1}{2\sqrt u}\,du.$

Integrating:

$\int\dfrac{1}{2\sqrt u}\,du=\sqrt u+C=\sqrt{\tan x}+C.$

Put $u = 1 - x$, so $du = -dx$; the integral becomes $-\int u^{-1/2}\,du = -2\sqrt{u} = -2\sqrt{1-x} + c$.

Let $u = 1 - x \Rightarrow du = -dx \Rightarrow dx = -du$. Then

$\int \frac{dx}{\sqrt{1 - x}} = \int \frac{-du}{\sqrt{u}} = -\int u^{-1/2}\,du = -\,\frac{u^{1/2}}{\tfrac12} = -2\sqrt{u}.$

Substitute $u = 1 - x$ back:

$= -2\sqrt{1 - x} + c.$

$\int (2x+4)^5\,dx=\dfrac{(2x+4)^6}{12}+c$.

Use the standard result $\int (ax+b)^n\,dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c$ (the extra $\tfrac1a$ accounts for the inner derivative). With $a=2$, $b=4$, $n=5$:

$\int (2x+4)^5\,dx=\frac{(2x+4)^{6}}{2\times6}+c=\frac{(2x+4)^6}{12}+c.$

The substitution $u=\sqrt x$ makes $du = \dfrac{dx}{2\sqrt x}$, converting the integral directly to $2\int \sin u\,du$.

Let $u=\sqrt x$. Then $du = \dfrac{1}{2\sqrt x}\,dx$, so $dx = 2\sqrt x\,du = 2u\,du$.

$\int \frac{\sin\sqrt x}{\sqrt x}\,dx = \int \frac{\sin u}{u}\cdot 2u\,du = 2\int \sin u\,du = -2\cos u + c$

Substituting back $u=\sqrt x$:

$= -2\cos\sqrt x + c$

Substituting $u = 1-x$ (or differentiating the given answer) confirms $\int (1-x)^{-2}\,dx = (1-x)^{-1} + c$, so the statement is True.

Evaluate the integral by substitution. Let $u = 1 - x$, so $du = -dx$, i.e. $dx = -du$:

$\int (1-x)^{-2}\,dx = \int u^{-2}(-du) = -\int u^{-2}\,du = -\left(\frac{u^{-1}}{-1}\right) = u^{-1} + c = (1-x)^{-1} + c.$

As a cross-check, differentiate the proposed answer:

$\frac{d}{dx}\big[(1-x)^{-1}\big] = -1\,(1-x)^{-2}\cdot(-1) = (1-x)^{-2},$

which is exactly the integrand. Both methods agree.

Put $u = 1 + e^x$; then $du = e^x\,dx$ matches the numerator exactly, giving $\int u^{-1/2}\,du = 2\sqrt{1+e^x} + C$.

Step 1 — Substitution. Let $u = 1 + e^x$. Then $\dfrac{du}{dx} = e^x$, i.e. $du = e^x\,dx$.

Step 2 — Rewrite the integral.

$\int \frac{e^x}{\sqrt{1 + e^x}}\,dx = \int \frac{du}{\sqrt{u}} = \int u^{-1/2}\,du.$

Step 3 — Integrate.

$\int u^{-1/2}\,du = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C = 2\sqrt{1 + e^x} + C.$