Let \( u^2 = 2x + 1 \Rightarrow u = \sqrt{2x+1} \).

Then \( 2u\,du = 2\,dx \Rightarrow du = dx/u \) and

\[ 6x = 3(2x+1) - 3 = 3u^2 - 3. \]

When \( x = 0 \), \( u = 1 \); when \( x = 1 \), \( u = \sqrt{3} \).

So

\[ \int_0^1 \frac{6x}{\sqrt{2x+1}}dx = \int_1^{\sqrt{3}} (3u^2 - 3)\,du = \bigl[u^3 - 3u\bigr]_1^{\sqrt{3}} = (3\sqrt{3} - 3\sqrt{3}) - (1 - 3) = 2. \]

Answer (3 s.f.): 2.00.

Let \( u = 1 + x^3 \Rightarrow du = 3x^2 dx \), so \( x^2dx = \tfrac{1}{3}du \).

When \( x = 0 \), \( u = 1 \); when \( x = 2 \), \( u = 9 \).

\[ \int_0^2 x^2\sqrt{1+x^3}\,dx = \int_1^9 \tfrac{1}{3}\sqrt{u}\,du = \tfrac{1}{3}\int_1^9 u^{1/2}du = \tfrac{1}{3}\Bigl[\tfrac{2}{3}u^{3/2}\Bigr]_1^9 = \tfrac{2}{9}(27 - 1) = \tfrac{52}{9}. \]

\( \tfrac{52}{9} \approx 5.78 \) (3 s.f.).

Answer: 5.78.

Use the trig substitution \( x = 3\sin\theta \).

Then \( dx = 3\cos\theta\,d\theta \) and

\[ \sqrt{9 - x^2} = \sqrt{9 - 9\sin^2\theta} = 3\cos\theta. \]

When \( x = 0 \), \( \theta = 0 \); when \( x = \tfrac{3}{2} \), \( \sin\theta = \tfrac{1}{2} \Rightarrow \theta = \tfrac{\pi}{6}. \)

So

\[ \int_0^{3/2} \frac{1}{\sqrt{9 - x^2}}\,dx = \int_0^{\pi/6} d\theta = \Bigl[\theta\Bigr]_0^{\pi/6} = \frac{\pi}{6}. \]

\( \dfrac{\pi}{6} \approx 0.524 \) (3 s.f.).

Answer: 0.524.

Let \( u = 1 + 2\sin x \Rightarrow du = 2\cos x\,dx \), so \( \cos x\,dx = \tfrac{1}{2}du \).

When \( x = 0 \), \( u = 1 \); when \( x = \tfrac{\pi}{6} \), \( u = 1 + 2\cdot\tfrac{1}{2} = 2 \).

\[ \int_0^{\pi/6} \frac{\cos x}{\sqrt{1+2\sin x}}dx = \int_1^2 \tfrac{1}{2}u^{-1/2}du = \tfrac{1}{2}\Bigl[2u^{1/2}\Bigr]_1^2 = [\sqrt{u}]_1^2 = \sqrt{2} - 1. \]

\( \sqrt{2} - 1 \approx 0.414 \) (3 s.f.).

Answer: 0.414.

Note that \( 2x + x^2 = x^2 + 2x \) and \( \dfrac{d}{dx}(x^2 + 2x) = 2x + 2 = 2(1+x) \).

Let \( u = x^2 + 2x \Rightarrow du = 2(1+x)dx \), so \( (1+x)dx = \tfrac{1}{2}du \).

When \( x = 0 \), \( u = 0 \); when \( x = 1 \), \( u = 3 \).

\[ \int_0^1 (1+x)\sqrt{2x+x^2}\,dx = \int_0^3 \tfrac{1}{2}\sqrt{u}\,du = \tfrac{1}{2}\int_0^3 u^{1/2}du = \tfrac{1}{2}\Bigl[\tfrac{2}{3}u^{3/2}\Bigr]_0^3 = \tfrac{1}{3}3^{3/2} = \sqrt{3}. \]

\( \sqrt{3} \approx 1.73 \) (3 s.f.).

Answer: 1.73.

Let \( u = e^{2x} + x \).

Then \( du = (2e^{2x} + 1)dx \), so the numerator is exactly \( du/dx \).

When \( x = 0 \), \( u = 1 \); when \( x = 1 \), \( u = 1 + e^2 \).

\[ \int_0^1 \frac{2e^{2x} + 1}{e^{2x} + x}\,dx = \int_1^{1+e^2} \frac{1}{u}\,du = [\ln|u|]_1^{1+e^2} = \ln(1+e^2). \]

Numerically, \( \ln(1+e^2) \approx 2.13 \) (3 s.f.).

Answer: 2.13.

Let \( u = \tan x \Rightarrow du = \sec^2 x\,dx \).

When \( x = 0 \), \( u = 0 \); when \( x = \tfrac{\pi}{3} \), \( u = \sqrt{3} \).

\[ \int_0^{\pi/3} \sec^2 x\tan^5 x\,dx = \int_0^{\sqrt{3}} u^5\,du = \Bigl[\tfrac{1}{6}u^6\Bigr]_0^{\sqrt{3}} = \tfrac{1}{6}(\sqrt{3})^6 = \tfrac{27}{6} = 4.5. \]

Answer (3 s.f.): 4.50.

Let \( u = \cos(x^2) \Rightarrow du = -2x\sin(x^2)dx \), so \( -2\,du = 4x\sin(x^2)dx \).

When \( x = 0 \), \( u = \cos 0 = 1 \); when \( x = \sqrt{\pi/3} \), \( u = \cos(\pi/3) = \tfrac{1}{2} \).

\[ \int_0^{\sqrt{\pi/3}} 4x\sin(x^2)dx = \int_1^{1/2} -2\,du = \Bigl[-2u\Bigr]_1^{1/2} = -2\cdot\tfrac{1}{2} - (-2\cdot 1) = -1 + 2 = 1. \]

Answer (3 s.f.): 1.00.

Let \( x = u^2 \Rightarrow dx = 2u\,du \).

Then \( x - \sqrt{x} = u^2 - u = u(u-1) \), so

\[ \frac{1}{x - \sqrt{x}}dx = \frac{1}{u^2-u} 2u\,du = \frac{2}{u-1}\,du. \]

When \( x = 2 \), \( u = \sqrt{2} \); when \( x = 3 \), \( u = \sqrt{3} \).

\[ \int_2^3 \frac{1}{x-\sqrt{x}}dx = \int_{\sqrt{2}}^{\sqrt{3}} \frac{2}{u-1}\,du = 2[\ln|u-1|]_{\sqrt{2}}^{\sqrt{3}} = 2\ln(\sqrt{3}-1) - 2\ln(\sqrt{2}-1). \]

Numerically this is approximately \( 1.14 \) (3 s.f.).

Answer: 1.14.

We need to evaluate

\[ \int_1^e \frac{(\ln x)^4}{x}\,dx. \]

Let \( u = \ln x \Rightarrow du = \dfrac{1}{x}dx \).

When \( x = 1 \), \( u = 0 \); when \( x = e \), \( u = 1 \).

\[ \int_1^e \frac{(\ln x)^4}{x}dx = \int_0^1 u^4\,du = \Bigl[\tfrac{1}{5}u^5\Bigr]_0^1 = \tfrac{1}{5}. \]

\( \tfrac{1}{5} = 0.200 \) (3 s.f.).

Answer: 0.200.

Let \( u = 2x + 1 \Rightarrow du = 2dx \), so \( dx = \tfrac{1}{2}du \) and \( x = \tfrac{u-1}{2} \).

Then

\[ \frac{x}{\sqrt{2x+1}}dx = \frac{\tfrac{u-1}{2}}{\sqrt{u}}\cdot\tfrac{1}{2}du = \tfrac{1}{4}(u^{1/2} - u^{-1/2})du. \]

When \( x = 0 \), \( u = 1 \); when \( x = 2 \), \( u = 5 \).

\[ \int_0^2 \frac{x}{\sqrt{2x+1}}dx = \int_1^5 \tfrac{1}{4}(u^{1/2} - u^{-1/2})du = \Bigl[\tfrac{1}{6}u^{3/2} - \tfrac{1}{2}u^{1/2}\Bigr]_1^5. \]

That simplifies to

\[ \frac{1}{3}\sqrt{5} + \frac{1}{3}. \]

Numerically, this is approximately \( 1.08 \) (3 s.f.).

Answer: 1.08.